Chapter 7 Chapter 7 Maintaining Mathematical Proficiency (p. 335)
12. 30 = 2
77 = 7
1. 3x − 7 + 2x = 3x + 2x − 7
= (3 + 2)x − 7
⋅3⋅5 ⋅ 11
The numbers have no common prime factors. So, the GCF of 30 and 77 is 1.
= 5x − 7 2. 4r + 6 − 9r − 1 = 4r − 9r + 6 − 1
= (4 − 9)r + 6 − 1
13. It is not possible for two integers to have no common factors
because 1 is a factor of every integer.
= −5r + 5
Chapter 7 Mathematical Practices (p. 356)
3. −5t + 3 − t − 4 + 8t = −5t − t + 8t + 3 − 4
1. The algebraic expression is 3x 2 − x + 1.
= (−5 − 1 + 8)t + 3 − 4 = (2)t + (−1) = 2t − 1 = 3s − 3 + 5
5. The algebraic expression is x 2 + 2.
= 3s + 2 5. 2m − 7(3 − m) = 2m − 7(3) − 7(−m)
= 2m − 21 + 7m = 9m − 21 = 4h + 24 − h + 2 = 4h − h + 24 + 2 = (4 − 1)h + 24 + 2 = 3h + 26 7. 20 = 2
⋅3⋅7 63 = 3 ⋅ 3 ⋅ 7
⋅ 2 = 4.
8. 42 = 2
The GCF of 42 and 63 is 3
⋅3⋅3⋅3 81 = 3 ⋅ 3 ⋅ 3 ⋅ 3
⋅
7 = 21.
9. 54 = 2
The GCF of 54 and 81 is 3
⋅2⋅2⋅3⋅3 84 = 2 ⋅ 2 ⋅ 3 ⋅ 7
⋅ ⋅ 3
3 = 27.
2
9. The algebraic expression is 2x 2.
7.1 Explorations (p. 357) 1. Step 1
(3x + 2) + (x − 5)
Step 2
4x + 2 − 5
Step 3
4x + (2 − 2) − 3
Step 4
4x − 3
Step 2
( x 2 + 2x + 2 ) − (x − 1) ( x 2 + 2x + 2 ) + (−x + 1)
Step 3
x 2 + 2x − x + 3
Step 4
x 2 + x + (x − x) + 3
Step 5
x2 + x + 3
2. Step 1
the opposite of each of its terms. Use the Commutative and Associative Properties to rearrange the terms so that like terms are grouped together. Then add or subtract like terms. 4. a. ( x 2 + 2x − 1 ) + ( 2x 2 − 2x + 1 )
⋅ 2 ⋅ 3 = 12.
⋅ ⋅ 64 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 The GCF of 28 and 64 is 2 ⋅ 2 = 4
11. 28 = 2
7. The algebraic expression is −x 2 + 2x.
3. Sample answer: If a polynomial is being subtracted, find
10. 72 = 2
The GCF of 72 and 84 is 2
6. The algebraic expression is x − 6.
8. The expression is 9.
6. 4(h + 6) − (h − 2) = 4(h) + 4(6) − h − (−2)
The GCF of 20 and 36 is 2
3. The algebraic expression is −x 2 − 2x. 4. The algebraic expression is x 2 + x − 1.
4. 3(s − 1) + 5 = 3(s) − 3(1) + 5
⋅2⋅5 36 = 2 ⋅ 2 ⋅ 3 ⋅ 3
2. The algebraic expression is −x 2 + x.
7
= x 2 + 2x 2 + 2x − 2x − 1 + 1 = ( x 2 + 2x2 ) + (2x − 2x) + (−1 + 1) = 3x 2 + 0 + 0 = 3x 2 b. (4x + 3) + (x − 2) = 4x + x + 3 − 2
= (4x + x) + (3 − 2) = 5x + 1
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Algebra 1 Worked-Out Solutions
379
Chapter 7 c. ( x 2 + 2 ) − ( 3x 2 + 2x + 5 )
9.
= ( x 2 + 2 ) + ( −3x 2 − 2x − 5 ) =
x2
−
3x 2
+ 7x 2 − x
− 2x + 2 − 5
8x 2 − 2x − 2
= ( 1x 2 − 3x 2 ) − 2x + (2 − 5) =
−2x 2
x2 − x − 2
The sum is 8x 2 − 2x − 2.
− 2x − 3
d. (2x − 3x) − ( x 2 − 2x + 4 ) = (−x) + ( −x 2 + 2x − 4 )
10. ( p 2 + p + 3 ) − ( −4p 2 − p + 3 )
= p 2 + p + 3 + 4p 2 + p − 3
= −x − x 2 + 2x − 4
= p 2 + 4p 2 + p + p + 3 − 3
= −x 2 − x + 2x − 4
= ( p 2 + 4p 2 ) + ( p + p) + (3 − 3)
= −x 2 + (−x + 2x) − 4
= 5p 2 + 2p + 0
= −x 2 + x − 4
= 5p 2 + 2p The difference is 5p 2 + 2p.
7.1 Monitoring Progress (pp. 358–361) 1. In the monomial −3x 4, the exponent of x is 4. So, the degree
of the monomial is 4.
−k + 5
11.
−( 3k 2
5
3. In the monomial —3 y, the exponent of y is 1. So, the degree of
− 6) ⇒
+6
−3k 2 − k + 11
2. In the monomial 7c 3d 2, the exponent of c is 3, and the
exponent of d is 2. So, the degree of the monomial is 3 + 2, or 5.
−k+5 −3k 2
The difference is −3k 2 − k + 11. 12. a. Penny:
−16t2 − 25t + 200
Paintbrush: − ( −16t2
+ 100 ) ⇒ +16t2
the monomial is 1.
− 100 −25t + 100
4. You can rewrite −20.5 as
−20.5x 0.
So, the degree of the
monomial is 0. 5. You can write the polynomial 4 − 9z in standard form as
−9z + 4.
The greatest degree is 1, so the degree of the polynomial is 1. The leading coefficient is −9. The polynomial has 2 terms, so it is a binomial. 6. You can write the polynomial t 2 − t 3 − 10 t in standard form
as −t 3 + t 2 − 10 t.
The greatest degree is 3, so the degree of the polynomial is 3. The leading coefficient is −1. The polynomial has 3 terms, so it is a trinomial. 7. You can write the polynomial 2.8x + x 3 in standard form as
x 3 + 2.8x.
The greatest degree is 3, so the degree of the polynomial is 3. The leading coefficient is 1. The polynomial has 2 terms, so it is a binomial. 8. (b − 10) + (4b − 3) = b + 4b − 10 − 3
= (b + 4b) + (−10 − 3) = 5b + (−13) = 5b − 13 The sum is 5b − 13.
380
−16t2 − 25t + 200
Algebra 1 Worked-Out Solutions
The polynomial −25t + 100 represents the distance between the objects after t seconds. b. When t = 0, the distance between the objects is
−25(0) + 100 = 100 feet. So, the constant term 100 indicates the distance between the penny and the paintbrush is 100 feet when they begin to fall.
As the value of t increases by 1, the value of −25t + 100 decreases by 25. This means that the objects become 25 feet closer to each other after each second. So, the coefficient −25 of the linear term represents how much the distance between the objects changes each second. 7.1 Exercises (pp. 362–364) Vocabulary and Core Concept Check 1. A polynomial in one variable is in standard form when the
exponents of the terms decrease from left to right. 2. The polynomial must have three terms, and the highest
degree must be 5. Sample answer: One possible polynomial is 2x 5 − x 2 + 7. 3. To determine whether a set of numbers is closed under an
operation, determine if the operation performed on any two numbers in the set results in a number that is also in the set. 4. The expression that does not belong with the other three is
x 2 − 8 x. This expression has a variable in one of its exponents, so it is not a polynomial. The other three expressions are polynomials because the terms are constants or other monomials whose variables have only whole number exponents.
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Chapter 7 Monitoring Progress and Modeling with Mathematics 5. In the monomial 4g, the exponent of g is 1. So, the degree of
the monomial is 1.
form as 3z 4 + 2z 3 + 5z.
The greatest degree is 4, so the degree of the polynomial is 4.
6. In the monomial 23x 4, the exponent of x is 4. So, the degree
The leading coefficient is 3. The polynomial has 3 terms, so it is a trinomial.
of the monomial is 4. 7. In the monomial −1.75k 2, the exponent of k is 2. So, the
degree of the monomial is 2. 4
18. You can write the polynomial 5z + 2z 3 + 3z 4 in standard
4
8. You can rewrite −—9 as −—9 x 0. So, the degree of the monomial
is 0. 9. In the monomial 7s 8t, the exponent of s is 8, and the exponent
of t is 1. So, the degree of the monomial is 8 + 1, or 9. 10. In the monomial 8m2n4, the exponent of m is 2, and the exponent
of n is 4. So, the degree of the monomial is 2 + 4, or 6. 11. In the monomial 9x y 3z7, the exponent of x is 1, the exponent
of y is 3, and the exponent of z is 7. So, the degree of the monomial is 1 + 3 + 7, or 11. 12. In the monomial −3q 4rs6, the exponent of q is 4, the
exponent of r is 1, and the exponent of s is 6. So, the degree of the monomial is 4 + 1 + 6, or 11. 13. You can write the polynomial 6c 2 + 2c 4 − c in standard
form as 2c 4 + 6c 2 − c.
5
19. You can write the polynomial πr 2 − —7 r 8 + 2r 5 in standard 5 form as −—7 r 8 + 2r 5 + πr 2.
The greatest degree is 8, so the degree of the polynomial is 8. 5
The leading coefficient is −—7. The polynomial has 3 terms, so it is a trinomial. —
20. The polynomial √ 7 n 4 is in standard form.
The only term has a degree of 4, so the degree of the polynomial is 4. —
The leading coefficient is √7 . The polynomial has 1 term, so it is a monomial. 4
21. The expression —3 πr 3 is a monomial because it is the product of a number, —43 π, and a variable with a whole number
exponent, r 3. The only variable has an exponent of 3, so the degree of the monomial is 3.
22. The expression 400x 8 + 600x 6 has two terms, so it is a
binomial. The greatest degree is 8, so the degree of the binomial is 8.
The greatest degree is 4, so the degree of the polynomial is 4. The leading coefficient is 2.
23. (5y + 4) + (−2y + 6) = 5y − 2y + 4 + 6
= (5y − 2y) + (4 + 6)
The polynomial has 3 terms, so it is a trinomial. 14. You can write the polynomial 4w11 − w12 in standard form
as −w12 + 4w11.
= 3y + 10 24. (−8x − 12) + (9x + 4) = −8x + 9x − 12 + 4
The greatest degree is 12, so the degree of the polynomial is 12.
= (−8x + 9x) + (−12 + 4)
The leading coefficient is −1.
= x + (−8) =x−8
The polynomial has 2 terms, so it is a binomial. 15. You can write the polynomial 7 + 3p 2 in standard form as
3p 2 + 7.
The greatest degree is 2, so the degree of the polynomial is 2. The leading coefficient is 3. The polynomial has 2 terms, so it is a binomial. 16. You can write the polynomial 8d − 2 − 4d 3 in standard form
as
−4d 3
+ 8d − 2.
The greatest degree is 3, so the degree of the polynomial 3.
Alternate solution: −8x − 12 + 9x + 4 x− 8 25. ( 2n2 − 5n − 6 ) + ( −n2 − 3n + 11 )
= 2n2 − n2 − 5n − 3n − 6 + 11 = ( 2n2 − n2 ) + (−5n − 3n) + (−6 + 11) = n2 − 8n + 5
The leading coefficient is −4. The polynomial has 3 terms, so it is a trinomial. 17. The polynomial 3t 8 is in standard form.
The only term has a degree of 8, so the degree of the polynomial is 8. The leading coefficient is 3. The polynomial has 1 term, so it is a monomial. Copyright © Big Ideas Learning, LLC All rights reserved.
Algebra 1 Worked-Out Solutions
381
Chapter 7 ( −3p 3 + 5p 2 − 2p ) + ( −p 3 − 8p 2 − 15p )
26.
= −3p 3 − p 3 + 5p 2 − 8p 2 − 2p − 15p =(
) + ( − ) + (−2p − 15p) = −4p 3 + ( −3p 2 ) + (−17p) =
−3p 3
−4p 3
−
−
p3
5p 2
8p 2
− 17p
3p 2
= 7m2 − 11m − 2
−4p 3 − 3p 2 − 17p
(
− g) + (
= −2y 2 + 2y + 18
= ( 4m2 + 3m2 ) + (−m − 10m) + (2 − 4)
+ −p 3 − 8p 2 − 15p
27.
= ( y 2 − 3y 2 ) + (−4y + 6y) + (9 + 9)
= 4m2 − m + 2 + 3m2 − 10m − 4
−3p 3 + 5p 2 − 2p
3g2
= y 2 − 4y + 9 − 3y 2 + 6y + 9
34. ( 4m2 − m + 2 ) − ( −3m2 + 10m + 4 )
Alternate solution:
3g2
33. ( y 2 − 4y + 9 ) − ( 3y 2 − 6y − 9 )
Alternate solution:
− 8g + 4 )
= 3g2 + 3g2 − g − 8g + 4
4m2 −
m+2
4m2 −
= ( 3g2 + 3g2 ) + (−g − 8g) + 4 = 6g2 + (−9g) + 4
28.
7m2 − 11m − 2 35. ( k 3 − 7k + 2 ) − ( k 2 − 12 ) = k 3 − 7k + 2 − k 2 + 12
= 6g2 − 9g + 4
( 9r 2
= k 3 − k 2 − 7k + (2 + 12)
+ 4r − 7 ) +
( 3r 2
− 3r )
= 9r 2 + 3r 2 + 4r − 3r − 7 = ( 9r 2 + 3r 2 ) + (4r − 3r) − 7 = 12r 2 + r − 7
= k 3 − k 2 − 7k + 14 36. (−r − 10) − ( −4r 3 + r 2 + 7r )
= −r − 10 + 4r 3 − r 2 − 7r = 4r 3 − r 2 + (−r − 7r) − 10
Alternate solution:
= 4r 3 − r 2 − 8r − 10
9r 2 + 4r − 7
Alternate solution:
+ 3r 2 − 3r
−r − 10
12r 2 + r − 7
−
( −4r 3
+
r2
−r − 10
+ 7r ) ⇒ +
29. ( 4a − a3 − 3 ) + ( 2a3 − 5a2 + 8 )
= −a3 + 2a3 − 5a2 + 4a − 3 + 8 = ( −a3 + 2a3 ) − 5a2 + 4a (−3 + 8) = a3 − 5a2 + 4a + 5 30.
( s3
− 2s − 9 ) +
( 2s 2
=
−
6s 3 )
+
2s 2
−
+ s)
+ (−2s + s) − 9
= −5s 3 + 2s 2 − s − 9
+
−6s 3
+
− 7r
37. ( t 4 − t 2 + t ) − ( 12 − 9t 2 − 7t )
= t 4 − t 2 + t − 12 + 9t 2 + 7t = t 4 + 8t 2 + 8t − 12 Alternate solution: t4 − t2 + t
t4 − t2 + t
− ( −9t 2 − 7t + 12 ) ⇒ + + 9t 2 + 7t − 12 t 4 + 8t 2 + 8t − 12
− 2s − 9 2s 2
−
r2
4r 3 − r 2 − 8r − 10
Alternate solution: s3
4r 3
= t 4 + ( −t 2 + 9t 2 ) + (t + 7t) − 12 6s 3
= s 3 − 6s 3 + 2s 2 − 2s + s − 9
( s3
m+2
− ( −3m2 + 10m + 4 ) ⇒ 3m2 − 10m − 4
+s
−5s 3 + 2s 2 − s − 9 31. (d − 9) − (3d − 1) = d − 9 − 3d + 1
= (d − 3d) + (−9 + 1) = −2d − 8 32. (6x + 9) − (7x + 1) = 6x + 9 − 7x − 1
= (6x − 7x) + (9 − 1) = −x + 8
38. ( 4d − 6d 3 + 3d 2 ) − (10d 3 + 7d − 2)
= 4d − 6d 3 + 3d 2 − 10d 3 − 7d + 2 = ( −6d 3 − 10d 3 ) + 3d 2 + (4d − 7d) + 2 = −16d 3 + 3d 2 − 3d + 2 39. When writing the subtraction as addition, the last term of the
polynomial was not multiplied by −1.
( x2 + x ) − ( 2x2 − 3x ) = x2 + x − 2x2 + 3x = ( x2 − 2x2 ) + (x + 3x) = −x2 + 4x
Alternate solution: 6x + 9 6x + 9 − (7x + 1) ⇒ −7x − 1 −x + 8
382
Algebra 1 Worked-Out Solutions
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Chapter 7 40. The terms −4x2 and 8x are not like terms, so they cannot be
51. −16t 2 + v0t + s0 = −16t 2 + (−45)t + 200
added. x3 − 4x2 + −3x3
= −16t 2 − 45t + 200
+3
Let t = 1.
+ 8x − 2
−16t 2 − 45t + 200 = −16(1)2 − 45(1) + 200
−2x3 − 4x2 + 8x + 1
= −16 − 45 + 200 = 139
41. (8b + 6) − (4 + 5b) = 8b + 6 − 4 − 5b
= (8b − 5b) + (6 − 4) = 3b + 2 A polynomial that represents how much more it costs to make b necklaces than b bracelets is 3b + 2. 42. (142 + 12m) + (52 + 6m) = 142 + 52 + 12m + 6m
= (142 + 52) + (12m + 6m)
A polynomial that represents the height of the water balloon is −16t 2 − 45t + 200, and the water balloon is 139 feet from the ground after 1 second. 52. −16t 2 + v0t + s0 = −16t 2 + 16t + 3
Let t = 1. −16t 2 + 16t + 3 = −16(1)2 + 16(1) + 3 = −16(1) + 16 + 3
= 194 + 18m A polynomial that represents the total number of memberships at the fitness center is 194 + 18m. 2s2 − 5st − t 2
43.
−
( s2
+ 7st −
t2 )
2s2 − 5st − t 2 ⇒ + −s2 − 7st + t 2 s2 − 12st
44. ( a2 − 3ab + 2b2 ) + ( −4a2 + 5ab − b2
)
= −16 + 16 + 3 =3 The polynomial −16t 2 + 16t + 3 represents the height of the tennis ball after t seconds, and the tennis ball is 3 feet high after 1 second. So, it is back to the initial height where you hit it with the racket. 53. a. ( −16t 2 + 98 ) − ( −16t 2 + 46t + 6 )
= −16t 2 + 98 + 16t 2 − 46t − 6
= a2 − 4a2 − 3ab + 5ab + 2b2 − b2
= ( −16t 2 + 16t 2 ) − 46t + (98 − 6)
= ( a2 − 4a2 ) + (−3ab + 5ab) + ( 2b2 − b2 )
= −46t + 92
= −3a2 + 2ab + b2 45.
− 6d 2
The polynomial −46t + 92 represents the distance between the objects after t seconds.
+ c 2 − 2cd + 2d 2
b. When t = 0, the distance between the objects is
c2
−46(0) + 92 = 92 feet. So, the constant term 92 indicates that the distance between the two balls is 92 feet when they are thrown.
2c 2 − 2cd − 4d 2 46. ( −x 2 + 9xy ) − ( x 2 + 6xy − 8y 2 )
= −x 2 + 9xy − x 2 − 6xy + 8y 2
As the value of t increases by 1, the value of −46t + 92 decreases by 46. This means that the two balls become 46 feet closer to each other after each second. So, the coefficient, −46, of the linear term represents how much the distance between objects changes each second.
= ( −x2 − x2 ) + (9xy − 6xy) + 8y 2 = −2x 2 + 3xy + 8y 2 47. The terms of a polynomial are always monomials.
A polynomial is a monomial or a sum of monomials, and each monomial is a term of the polynomial. 48. The difference of two trinomials is sometimes a trinomial. If
like terms have the same coefficient, they will cancel when subtracted, so the difference will have fewer than 3 terms. Or, if the terms in the trinomial are not all of the same degree, then the difference could have more than 3 terms. 49. A binomial is sometimes a polynomial of degree 2. The two
terms in the binomial can be of any degree. 50. The sum of two polynomials is always a polynomial.
Polynomials are closed under addition.
−0.028t 3 + 0.06t 2 + 0.1t + 17
54. a.
+
− 0.38t 2 + 1.5t + 42 −0.028t 3 − 0.32t 2 + 1.6t + 59
The polynomial −0.028t 3 − 0.32t 2 + 1.6t + 59 represents the total amount spent each year on buying new and used vehicles in the 7-year period. b. Let t = 5.
−0.028t 3 − 0.32t 2 + 1.6t + 59 = −0.028(5)3 − 0.32(5)2 + 1.6(5) + 59 = −0.028(125) − 0.32(25) + 8 + 59 = −3.5 − 8 + 8 + 59 = 55.5 The total amount spent on buying new and used vehicles in the fifth year was $55.5 million.
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Algebra 1 Worked-Out Solutions
383
Chapter 7 b. 2x2 − 14x = 2(20)2 − 14(20)
55. P = 2x + (3x − 2) + (2x + 1) + (5x − 2)
= 2(400) − 280
= (2x + 3x + 2x + 5x) + (−2 + 1 − 2)
= 800 − 280
= 12x − 3
= 520
The polynomial 12x − 3 represents the perimeter of the quadrilateral. 56. a. −2x + 1 − (x − 2)
x − 2 − (−2x + 1)
or
= −2x + 1 − x + 2
= x − 2 + 2x − 1
= (−2x − x) + (1 + 2)
= (x + 2x) + (−2 − 1)
= −3x + 3
−3
0 −3
+3 +3
−3x = −3 −3x −3 —=— −3 −3 x=1
3x = 3 3x 3 —=— 3 3 x=1
a sum of 1
—2 x 2 +
+ —12 x 2 −
−
(
1 —2 x 2 1 —2 x 2
+ −
1 —2 1 —2
)
Area of Area of Area of Area of patio = rectangular area − pool − hot tub =
ℓw
−
ℓw
−
s2
= 8x(4x) − 10(4x) − (12x2 − 28x) − x2 = 32x2 − 40x − 12x2 + 28x − x2 = ( 32x2 − 12x2 − x2 ) + (−40x + 28x) = 19x2 − 12x b. 19x2 − 12x = 19(9)2 − 12(9)
1 —2
+ and
1 —2 x 2
1 —2
− have
and a difference of 1.
1 —2 1 —2
61. a.
= (8x − 10)(4x) − ( 6x(2x) − 14(2x) ) − x2
associative, so you can add in any order.
x2
So, you need 1.3 gallons of sealant to cover the deck once.
A = (8x − 10)(2x + 2x) − (6x − 14)(2x) − x2
57. Your friend is correct. Addition is commutative and 1 —2 x 2
= 1.3 — 400
A
Each expression equals 0 when x = 1, which is the x-coordinate of the point of intersection of the two lines.
58. Sample answer: The polynomials
covered by 1 gallon of sealant to get the amount of sealant needed. 520
3x − 3 = 0
or
c. Divide the total area of the deck by the area that can be
= 3x − 3
or
The vertical distance between points on the lines with the same x-value can be represented by the absolute value of −3x + 3 or 3x − 3. b. −3x + 3 =
When x = 20 feet, the total area of the deck is 520 square feet.
= −
1 —2 x 2 1 —2 x 2
+ +
x2
1 —2 1 —2
1
59. a. The set of negative integers is not closed under
multiplication because the product of two negative integers is always a positive integer. b. The set of whole is closed under addition because the sum
of two whole numbers is always a whole number.
= 19(81) − 108 = 1539 − 108 = 1431 So, when x = 9, the area of the patio is 1431 square feet. The patio costs 1431(10) = $14,310. Maintaining Mathematical Proficiency 62. 2(x − 1) + 3(x + 2) = 2(x) − 2(1) + 3(x) + 3(2)
= 2x − 2 + 3x + 6 = (2x + 3x) + (−2 + 6) = 5x + 4 63. 8(4y − 3) + 2( y − 5) = 8(4y) − 8(3) + 2( y) − 2(5)
60. a. Level 1: A = ℓw
= x( 10 + (x − 12) ) = x( x + (10 − 12) ) = x(x − 2) = x(x) − x(2) = x 2 − 2x Level 2: A = ℓw = x(x − 12) = x(x) − x(12) = x 2 − 12x
= 32y − 24 + 2y − 10 = (32y + 2y) + (−24 − 10) = 34y − 34 64. 5(2r + 1) − 3(−4r + 2) = 5(2r) + 5(1) − 3(−4r) − 3(2)
= 10r + 5 + 12r − 6 = (10r + 12r) + (5 − 6) = 22r − 1
Total area:
( x 2 − 2x ) + ( x 2 − 12x ) = x 2 + x 2 − 2x − 12x = ( x 2 + x 2 ) + (−2x − 12x) = 2x 2 − 14x
384
Algebra 1 Worked-Out Solutions
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Chapter 7 7.2 Explorations (p. 365)
7.2 Monitoring Progress (pp. 366–368)
1. a. 1; The product of 1 and 1 is 1.
1. ( y + 4)( y + 1) = y( y + 1) + 4( y + 1)
b. −1; The product of 1 and −1 is −1.
= y( y) + y(1) + 4( y) + 4(1)
c. 1; The product of −1 and −1 is 1.
= y 2 + y + 4y + 4
d. x; The product of any number and 1 is that number.
= y 2 + 5y + 4
e. −x; The product of any number and −1 is the opposite of
The product is y2 + 5y + 4.
that number. f. −x; The product of any number and 1 is that number.
×
g. x; The product of the opposite of a number and −1 is the
i.
z+6 6z − 12
number. h.
z−2
2.
The product of any number multiplied by itself is the number squared.
z2 − 2z
−x2;
The product is z2 + 4z − 12.
x2;
The product of any number and its opposite is the opposite of the number squared.
j. x 2; The product of any number multiplied by itself is the
z2 + 4z − 12 3. ( p + 3)( p − 8) = ( p + 3)[ p + (−8) ]
number squared. 2. a. (x + 3)(x − 2) =
x2
+ (−2x + 3x) − 6
=
x2
+x−6
p −8
b. (2x − 1)(2x + 1) = 4x 2 + (2x − 2x) − 1
=
4x2
−1
c.
(x + 2)(2x − 1) =
2x2
d.
3 3p
−8p
−24
The product is
p2
+ 3p − 8p − 24, or p2 − 5p − 24.
4. (r − 5)(2r − 1) = [ r + (−5) ][ 2r + (−1) ]
+ (−x + 4x) − 2
= 2x2 + 3x − 2
p p2
r
−5
2r
2r2
−10r
−1
−r
5
The product is 2r 2 − 10r − r + 5, or 2r 2 − 11r + 5. 5.
First
Outer
Inner
Last
(m − 3)(m − 7) = m(m) + m(−7) + (−3)(m) + (−3)(−7) = m2 + (−7m) + (−3m) + 21 = m2 − 10m + 21 The product is m2 − 10m + 21. (−x − 2)(x − 3) = −x2 + (3x − 2x) + 6
6.
= −x2 + x + 6
First Outer
Inner
(x − 4)(x + 2) = x(x) + x(2) + (−4)(x) + (−4)(2) = x2 + 2x + (−4x) + (−8)
3. Multiply each term in one polynomial by each term in the
= x2 − 2x − 8
other polynomial, then combine like terms.
The product is x2 − 2x − 8.
4. Sample answer: 7.
First
Outer
Inner
( 2u + )( u − ) = 2u(u) + 2u ( ) 1 —2
3 −—2
3 —2
= 2u2 + (−3u) + (x + 1)(x − 2) = x2 + (−2x + x) − 2 = x2 − x − 2
Copyright © Big Ideas Learning, LLC All rights reserved.
Last
+ —12 (u) 1 3 —2 u − —4
Last +
1 —2
( −— ) 3 2
= 2u2 − —52u − —34 The product is 2u2 − —52 u − —34 .
Algebra 1 Worked-Out Solutions
385
Chapter 7 8.
First
Outer
Inner
Last
(n + 2)( n2 + 3 ) = n( n2 )+ n(3) + 2( n2 ) + 2(3)
4. ( y + 6)( y + 4) = y( y + 4) + 6( y + 4)
= y( y) + y(4) + 6( y) + 6(4)
= n3 + 3n + 2n2 + 6 =
n3
+
2n2
= y2 + 4y + 6y + 24
+ 3n + 6
= y2 + 10y + 24
The product is n3 + 2n2 + 3n + 6.
5. (z − 5)(z + 3) = z(z + 3) − 5(z + 3)
x2 + 5x + 8
9.
×
= z(z) + z(3) − 5(z) − 5(3)
x+1
= z2 + 3z − 5z − 15
x2 + 5x + 8 x3
+
5x2
= z2 − 2z − 15
+ 8x
6. (a + 8)(a − 3) = a(a − 3)+ 8(a − 3)
x3 + 6x2 + 13x + 8
= a(a) − a(3) + 8(a) − 8(3)
The product is x3 + 6x2 + 13x + 8. n2
10.
= a2 − 3a + 8a − 24
− 2n + 4
×
= a2 + 5a − 24
n−3
7. (g − 7)(g − 2) = g(g − 2) − 7(g − 2)
−3n2 + 6n − 12
= g(g) − g(2) − 7(g) − 7(−2)
n3 − 2n2 + 4n
= g2 − 2g − 7g + 14
n3 − 5n2 + 10n − 12
= g2 − 9g + 14
The product is n3 − 5n2 + 10n − 12.
8. (n − 6)(n − 4) = n(n − 4) − 6(n − 4)
1
1
11. —2 h ( b1 + b2 ) = —2 (x − 7)[ x(x + 11) ]
= n(n) − n(4) − 6(n) − 6(−4)
= —12 (x − 7)(2x + 11) F
O
I
= n2 − 4n − 6n + 24
L
= —12 [ 2x2 + 11x + (−14x) + (−77) ] = —12 ( 2x2 − 3x − 77 )
= n2 − 10n + 24 9. (3m + 1)(m + 9) = 3m(m + 9) + 1(m + 9)
= 3m(m) + 3m(9) + 1(m) + 1(9)
77 = x2 − —32 x − — 2
The polynomial that represents the area of the trapezoidal 77 region becomes x2 − —32 x − — . The longer base becomes 2 x + 11. Substituting this value in the formula for the area of a trapezoid along with the other unchanged values changes the linear and constant terms in the polynomial.
= 3m2 + 27m + m + 9 = 3m2 + 28m + 9 10. (5s + 6)(s − 2) = 5s(s − 2) + 6(s − 2)
= 5s(s) − 5s(2) + 6(s) − 6(2) = 5s2 − 10s + 6s − 12 = 5s2 − 4s − 12
7.2 Exercises (pp. 369–370) Vocabulary and Core Concept Check
11. (x + 3)(x + 2)
1. Sample answer: Distribute one of the binomials over each
term in the other binomial, and simplify by combining like terms. Or, write each binomial as a sum of terms and make a table of products. Then, write the sum of the products and simplify by combining like terms. 2. The letters stand for the sets of terms to multiply: first, outer,
x
3
x
x2
3x
2
2x
6
x2 + 3x + 2x + 6 = x2 + 5x + 6 12. ( y + 10)( y − 5) = ( y + 10)[ y + (−5) ]
inner, and last. Monitoring Progress and Modeling with Mathematics 3. (x + 1)(x + 3) = x(x + 3) + 1(x + 3)
= x(x) + x(3) + 1(x) + 1(3) = x2 + 3x + x + 3 = x2 + 4x + 3
386
Algebra 1 Worked-Out Solutions
y −5
y
10
y2
10y
−5y −50
y2 + 10y − 5y − 50 = y2 + 5y − 50
Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 13. (h − 8)(h − 9) = [ h + (−8) ][ h + (−9) ]
h −9
h
−8
h2
−8h
−9h
72
20. The two terms that represent x − 5 on the left side of the
table should be x and −5, not 5.
(x − 5)(3x + 1) = [ x + (−5) ](3x + 1)
x
h2 − 8h − 9h + 72 = h2 − 17h + 72
c −5
−6
c2
−6c
−5c
30
1
3x 2
x
−5 −15x −5
14. (c − 6)(c − 5) = [ c + (−6) ][ c + (−5) ]
c
3x
3x2 + x − 15x − 5 = 3x2 − 14x − 5 21.
First
= b2 + 7b + 3b + 21 = b2 + 10b + 21
c2 − 6c − 5c + 30 = c2 − 11c + 30 15. (3k − 1)(4k + 9) = [ 3k + (−1) ](4k + 9)
3k
−1
4k
12k2
−4k
9
27k
−9
22.
First
g
5g2
3g
8
40g
24
5g2 +
3g + 40g + 24 =
4j
−3
8j 2
−12j
−7 −14j
23.
First
Last
= k2 + 4k − 5 24.
First Outer
Inner
Last
(x − 4)(x + 8) = x(x) + x(8) + (−4)(x) + (−4)(8) 5g2
= x2 + 8x − 4x − 32
+ 43g + 24
= x2 + 4x − 32 25.
First Outer 1 —4
3 −—4
26.
3 4
1 4
First
Outer
Inner
Last
( z − — )( z − — ) = z(z) + z( −— ) + ( −— )(z) + ( −— )( −— ) 5 3
2 3
2 3
5 3
5 3
2 3
10 = z2 − —23 z − —53 z + — 9
84
10 = z2 − —73 z + — 9
15d 2 − 36d − 35d + 84 = 15d 2 − 71d + 84 27.
First
Outer
Inner
Last
(9 − r)(2 − 3r) = 9(2) + 9(−3r) + (−r)(2) + (−r)(−3r)
= t(t) + t(5) − 2(t) − 2(5)
= 18 − 27r − 2r + 3r 2
= t2 + 5t − 2t − 10
= 18 − 29r + 3r 2
= t2 + 3t − 10
= 3r 2 − 29r + 18
Copyright © Big Ideas Learning, LLC All rights reserved.
Last
3 = q2 − —12 q − — 16
−12
(t − 2)(t + 5) = t (t + 5) − 2(t + 5)
1 —4
3 = q2 + —14 q − —34 q − — 16
21
19. The first term t also should be multiplied by t + 5.
Inner
( q − )( q + ) = q(q) + q( ) + ( ) (q) + ( −— )( — ) 3 —4
15d 2 −36d
−7 −35d
Inner
= k2 − k + 5k − 5
18. (5d − 12)(−7 + 3d) = [ 5d + (−12) ][ 3d + (−7) ]
3d
Outer
(k + 5)(k − 1) = k(k) + k(−1) + 5(k) + 5(−1)
8j2 − 12j − 14j + 21 = 8j 2 − 26j + 21
5d
Last
= w2 + 15w + 54
17. (−3 + 2j )(4j − 7) = [ 2j + (−3) ][ 4j + (−7) ]
2j
Inner
= w2 + 6w + 9w + 54
16. (5g + 3)(g + 8)
3
Outer
(w + 9)(w + 6) = w(w) + w(6) + 9(w) + 9(6)
12k2 − 4k + 27k − 9 = 12k2 + 23k − 9
5g
Outer Inner Last
(b + 3)(b + 7) = b(b) + b(7) + 3(b) + 3(7)
Algebra 1 Worked-Out Solutions
387
Chapter 7 28.
First
Outer
Inner
Last
33.
(8 − 4x)(2x + 6) = 8(2x) + 8(6) + (−4x)(2x) + (−4x)(6) = 16x + 48 − 8x2 − 24x
Area of Area of Area of = − shaded region rectangular region triangular region A
= −8x2 + 16x − 24x + 48 = −8x2 − 8x + 48 29.
First
Outer
Inner
F
Last
(w + 5)( w2 + 3w ) = w( w2 ) + w(3w) + 5( w2 ) + 5(3w)
Outer
Inner
Last
(
=
L
= 2x(x) + 2x(5) + (−9)(x) + (−9)(5)
34.
+ 10x − 9x − 45
+
11 x — 2
− ℓw − (x − 7)(5)
1)2
= (x + 1)(x + 1) − (x − 7)(5) F
O
I
L
= x(x) + x(1) + 1(x) + 1(1) − [ x(5) − 7(5) ]
= —12 (p + 1)(2p − 6)
=
+ 5x + 6x + 30 )
)
= (x +
1
=
[ x(x) + x(5) + 6(x) + 6(5) ]
= s2
A
32. A = —2 bh
=
L
+ 15
The polynomial + x − 45 represents the area of the rectangular region.
1 =—2
I
+ 11x + 30 )
) (
1 —2 x 2
2x2
O
1 —2 ( x 2
1 —2 ( x 2
O
Area of Area of Area of = − shaded region square region rectangular region
= 2x2 + x − 45
F
F − —21
11 The polynomial —12 x2 + — x + 15 represents the area of the 2 shaded region.
= (2x − 9)(x + 5)
=
L
11 = x2 − —12 x2 + 11x − — x + (30 − 15) 2
31. A = ℓw
2x2
I
11 = x2 + 11x + 30 − —12 x2 + — x − 15 2
= v3 + 5v2 − 24v
I
O
= x2 + 11x + 30 − —12 ( x2 ) − —12 (11x) − —12 (30)
= v3 + 8v2 − 3v3 − 24v
O
− —12 (x + 6)(x + 5)
= (x2 + 11x + 30) −
(v − 3)( v2 + 8v ) = v( v2 ) + v(8v) + (−3)( v2 ) + (−3)(8v)
F
= (x + 6)(x + 5)
= ( x2 + 5x + 6x + 30 ) −
= w3 + 8w2 + 15w First
− —12 bh
= [ x(x) + x(5) + 6(x) + 6(5) ]
= w3 + 3w2 + 5w2 + 15w
30.
= ℓw
I
= x2 + x + x + 1 − (5x − 35)
L
[ p(2p) + p(−6) + 1(2p) + 1(−6) ] ( 2p2 − 6p + 2p − 6 ) ( 2p2 − 4p − 6 ) ( 2p2 ) − —12 (4p) − —12 (6)
1 —2 1 —2 1 —2
= p2 − 2p − 3
= x2 + 2x + 1 − 5x + 35 = x2 + 2x − 5x + 1 + 35 = x2 − 3x + 36 The polynomial x2 − 3x + 36 represents the area of the shaded region. x2 + 3x + 2
35.
×
The polynomial p2 − 2p − 3 represents the area of the triangular region.
x+4 4x2 + 12x + 8
x3 + 3x2 + 2x x3 + 7x2 + 14x + 8 f 2 + 4f + 8
36.
×
f+1 f 2 + 4f + 8
f 3 + 4f 2 + 8f f 3 + 5f 2 + 12f + 8 y2 + 8y − 2
37.
×
y+3 3y2 + 24y − 6
y3 + 8y2 − 2y y3 + 11y2 + 22y − 6
388
Algebra 1 Worked-Out Solutions
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Chapter 7 t2 − 5t + 1
38.
×
44. a. A = ℓw
t−2
= (x + 22 + x)(x + 20 + x)
−2t2 + 10t − 2 t3
−
5t2
= (2x + 22)(2x + 20)
+t
F
t3 − 7t2 + 11t − 2 ×
A polynomial that represents the combined area of the photo and the frame is ( 4x2 + 84x + 440 ) square inches.
20b2 + 20b − 16 −5b3 − 5b2 + 4b
b. Let x = 4.
−5b3 + 15b2 + 24b − 16 40.
4x2 + 84x + 440 = 4(4)2 + 84(4) + 440 = 4(16) + 336 + 440
−d+7
×
= 64 + 336 + 440
d+6
2d 3
−
d2
= 840
− 6d + 42
12d 2
+ 7d
When the width of the frame is 4 inches, the combined area of the photo and the frame is 840 square inches.
2d 3 + 11d 2 + d + 42
45. The degree of the product is the sum of the degrees of each
3e2 − 5e + 7
41.
×
binomial.
6e + 1
46. Sample answer: (2x − 6)( x 2 + 3x + 4 )
3e2 − 5e + 7 18e3 − 30e2 + 42e
x2 + 3x + 4
18e3 − 27e2 + 37e + 7 6v2
42.
×
2x − 6 −6x2
+ 2v − 9 −5v + 4
−
10v2
2x3
−10x − 24
The product is 2x3 − 10x − 24, which is a trinomial of degree 3.
+ 45v
−30v3 + 14v2 + 53v − 36
47. The FOIL method can only be used for multiplying two
43. a. A = ℓw
= (10x + 10)(4x + 20) F
− 18x − 24
2x3 + 6x2 + 8x
24v2 + 8v − 36 −30v3
L
= 4x2 + 84x + 440
−b + 4
2d2
I
= 4x2 + 40x + 44x + 440
5b2 + 5b − 4
39.
O
= 2x(2x) + 2x(20) + 22(2x) + 22(20)
O
I
L
=10x(4x) + 10x(20) + 10(4x) + 10(20) = 40x2 + 200x + 40x + 200 = 40x2 + 240x + 200 A polynomial that represents the area of the football field is ( 40x2 + 240x + 200 ) square feet.
binomials, because each of the four letters represent one of the products when two binomials are multiplied. When two trinomials are multiplied, however, there are 6 products. The FOIL method would leave out the products that include the middle terms of the two trinomials. 48. a. The two binomials being multiplied are (−4x + 3) and
(−8x − 9). The binomial (−4x + 3) is comprised of the two terms on top of the table, and the binomial (−8x − 9) is comprised of the two terms on the left side of the table.
b. When x > 0, a is positive because it is the product of two
b. 4x + 20 = 160
negative terms, b is negative because it is the product of one negative term and one positive term, c is positive because it is the product of two negative terms, and d is negative because it is the product of one negative term and one positive term.
− 20 − 20 4x = 140 4x 140 —=— 4 4 x = 35
49. Your answers should be equivalent. You are both multiplying
40x2 + 240x + 200 = 40(35)2 + 240(35) + 200
the same binomials, and neither the order in which you multiply nor the method used will make a difference.
= 40(1225) + 8400 + 200 = 49,000 + 8400 + 200 = 57,600 When the width of a football field is 160 feet, the area is 57,600 square feet. Copyright © Big Ideas Learning, LLC All rights reserved.
Algebra 1 Worked-Out Solutions
389
Chapter 7 54. y = −4 ∣ x + 2 ∣
50. V = ℓwh
= (4x − 3)(x + 1)(x + 2) = (4x − 3)[ x(x + 2) + 1(x + 2) ]
g(x) = −4[ −(x + 2) ],
= (4x −
g(x) = −4(−x − 2),
3)( x2
+ 2x + x + 2 )
if x + 2 < 0 if x + 2 ≥ 0
−4[ −(x + 2) ], {−4(x + 2),
= (4x − 3)[ x(x) + x(2) + 1(x) + 1(2) ]
if x + 2 < −2
= (4x − 3)( x2 + 3x + 2 )
g(x) = −4(−x) − 4(−2),
= 4x( x2 + 3x + 2 ) − 3( x2 + 3x + 2 )
g(x) = 4x + 8,
= 4x( x2 ) + 4x(3x) + 4x(2) − 3( x2 ) − 3(3x) − 3(2)
g(x) = −4(x + 2),
=
g(x) = −4(x) − 4(2),
−2
g(x) = −4x − 8,
if x ≥ −2
+
4x3
12x2
+ 8x −
3x2
− 9x − 6
= 4x3 + ( 12x2 − 3x2 ) + (8x − 9x) − 6 = 4x3 + 9x2 − x − 6 A polynomial that represents the volume of the container is ( 4x3 + 9x2 − x − 6 ) cubic feet. 51. a. The product of m and n is c. So, when c > 0, m and n have
the same signs because the product of two numbers with the same sign is positive. b. The product of m and n is c. So, when c < 0, m and n have
opposite signs because the product of one positive number and one negative number is negative. Maintaining Mathematical Proficiency 52. y = ∣ x ∣ + 4
A piecewise function for y = ∣ x ∣ + 4 is g(x) =
−x + 4, if x < 0 . x + 4, if x ≥ 0
{
g(x) =
6[ −(x − 3) ], { 6(x − 3),
g(x) = 6[ −(x − 3) ], g(x) = 6(−x + 3),
g(x) = 6(x − 3), g(x) = 6(x) − 6(3) g(x) = 6x − 18,
−2
if x < −2 if x + 2 ≥
0 −2
So, a piecewise function for y = −4 ∣ x + 2 ∣ is 4x + 8, if x < −2 . g(x) = −4x − 8, if x ≥ −2
{
55. 10 2
⋅ 10
⋅
9
= 10 2 + 9 = 1011
x5 + 1 x6 1 x x x 1 1 1 1 57. (3z 6)−3 = — =—=—=— (3z 6)3 33(z 6)3 27z 6 ⋅3 27z18 x5 x x
56. — =— = —8 = x 6 − 8 = x−2 = —2 8 8
( ) 2y 4 y
58. — 3
−2
1 1 = (2y 4 − 3)−2 = (2y)−2 = 2−2y−2 = — =— 22y2 4y2
7.3 Explorations (p. 371) 1. a. x 2 − 2x + 2x − 4 = x 2 − 4
2. a.
if x − 3 < 0 if x − 3 ≥ 0 if x − 3 <
0
+3 +3
g(x) = 6(−x) + 6(3), g(x) = −6x + 18,
0
b. 4x 2 + 2x − 2x − 1 = 4x 2 − 1
53. y = 6 ∣ x − 3 ∣
if x − 3 ≥ +3
x 2 + 2x + 2x + 4 = x 2 + 4x + 4 b.
if x < 3 0 +3
if x ≥ 3
So, a piecewise function for y = 6 ∣ x − 3 ∣ is −6x + 18, if x < 3 . g(x) = 6x − 18, if x ≥ 3
{
390
g(x) =
Algebra 1 Worked-Out Solutions
4x2 − 2x − 2x + 1 = 4x2 − 4x + 1
Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 3.
F
O
I
L
(a + b)(a − b) = a(a) + a(−b) + b(a) + b(−b) = a2 − ab + ab − b 2
(3x + 1)(3x − 1) = (3x)2 − 12 = 9x2 − 1
b2
+
+
+
−
+
+
+
+
+
+
+
+
+
+
+
+
−
Check
−
−
(a + b)(a − b) = a2 − b2
−
=
a2
c.
+
+
+
+
−
So, (a + b)(a − b) = a2 − b 2. (a + b)2 = (a + b)(a + b) F
O
I
L
= a(a) + a(b) + b(a) + b(b) = a2 + ab + ab + b2 = a2 + 2ab + b2 So, (a + b)2 = a2 + 2ab + b2. (a − b)2 = (a − b)(a − b) F
O
I
L
= a(a) +a(−b) + (−b)(a) + (−b)(−b) =
a2
− ab − ab +
So, (a −
=
a2
(x + 3)2 = x2 + 2(x)(3) + 32
− 2ab +
= x2 + 6x + 9
b2. Check
+
+ + +
+
+
+ + +
− − −
+ +
+ +
+ + + + + +
+
+
+ + +
4. a. ( a + b)( a − b) = a2 − b 2
(x + 3)(x − 3) =
x2
The product is 9x2 − 1. d. (a + b)2 = a2 + 2ab + b 2
b2
= a2 − 2ab + b2 b)2
9x2 − 3x + 3x − 1 = 9x2 − 1
−
32
= x2 − 9 Check
+ +
− −
+ + +
+ + +
− − − − − − − − −
−
+
x2 + 3x + 3x + 9 = x2 + 6x + 9 The product is x2 + 6x + 9. e. (a − b)2 = a2 − 2ab + b 2
(x − 2)2 = x2 − 2(x)(2) + 22
x2 − 3x + 3x − 9 = x2 − 9
= x2 − 4x + 4
The product is x2 − 9. −
− −
+
−
−
+ +
−
−
+ +
42
= x2 − 16 Check
+
− −
(x − 4)(x + 4) =
x2
+
Check
b. (a − b)(a + b) = a2 − b 2
+
+ + + +
+
+
+ + + +
− − − −
− − − −
− − − −
− − − −
− − − −
− − − −
x2 − 2x − 2x + 4 = x2 − 4x + 4 The product is x2 − 4x + 4.
x2 + 4x − 4x − 16 = x2 − 16 The product is x2 − 16.
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Algebra 1 Worked-Out Solutions
391
Chapter 7 9. a. The Punnett square shows four possible gene
f. (a + b)2 = a2 + 2ab + b 2
combinations of the offspring. Of these combinations, one results in black. So, —14 = 25% of the possible gene combinations result in black.
(3x + 1)2 = (3x)2 + 2(3x)(1) + 12 = 9x2 + 6x + 1 Check +
+
+
+
+
+
+
+
+
You can model the possible gene combinations of the offspring with the expression (0.5B + 0.5W )2. Notice that this product also represents the area of the Punnett square.
+
+
+
+
+
+
+
+
+
Expand the product to find the possible colors of the offspring.
+
+
+
+
+
(0.5B + 0.5W )2 = (0.5B)2 + 2(0.5B)(0.5W ) + (0.5W )2
The product is 9x2 + 6x + 1. 7.3 Monitoring Progress (pp. 372–374) 1. (x + 7)2 = x 2 + 2(x)(7) + 72
=
0.5B + 0.5W. There is an equal chance that the offspring inherits a black or a white gene from each parent.
+
9x2 + 3x + 3x + 1 = 9x2 + 6x + 1
x2
b. Model the gene from each parent with the expression
+ 14x + 49
The product is x2 + 14x + 49. 2. (7x − 3)2 = (7x)2 − 2(7x)(3) + 32
= 49x2 − 42x + 9 The product is 49x2 − 42x + 9. 3. (4x − y)2 = (4x)2 − 2(4x)(y) + y2
= 16x2 − 8xy + y2 The product is 16x2 − 8xy + y2. 4. (3m + n)2 = (3m)2 + 2(3m)(n) + n2
= 9m2 + 6mn + n2 The product is 9m2 + 6mn + n2. 5. (x + 10)(x − 10) = x 2 − 102
= x2 − 100 The product is x2 − 100. 6. (2x + 1)(2x − 1) = (2x)2 − 12
= 4x2 − 1 The product is 4x2 − 1. 7. (x + 3y)(x − 3y) = x 2 − (3y)2
= x2 − 9y2 The product is x2 − 9y2. 8. Rewrite 212 as (20 + 1)2. Then use the square of a binomial
pattern. (20 + 1)2 = 202 + 2(20)(1) + 12 = 400 + 40 + 1 = 441
= 0.25B2 + 0.5BW + 0.25W 2 Consider the coefficients in the polynomial. So, 25% of the offspring are BB, or black; 50% of the offspring are BW, or gray; and 25% of the offspring are WW, or white. 7.3 Exercises (pp. 375–376) Vocabulary and Core Concept Check 1. Substitute the first term of the binomial for a and the second
term of the binomial for b in the square of a binomial pattern a2 + 2ab + b2. Then simplify. 2. The expression that does not belong is (x + 2)(x − 3). It
is the only one that cannot be simplified using the sum and difference pattern. The pattern for this one is (a + b)(a − c), but the pattern for the others is (a + b)(a − b). Monitoring Progress and Modeling with Mathematics 3. (x + 8)2 = x 2 + 2(x)(8) + 82
= x2 + 16x + 64 4. (a − 6)2 = a2 − 2(a)(6) + 62
= a2 − 12a + 36 5. (2f − 1)2 = (2f )2 − 2(2f )(1) + 12
= 4f 2 − 4f + 1 6. (5p + 2)2 = (5p)2 + 2(5p)(2) + 22
= 25p2 + 20p + 4 7. (−7t + 4)2 = (−7t)2 + 2(−7t)(4) + 42
= 49t2 − 56t + 16 8. (−12 − n)2 = (−12)2 − 2(−12)(n) + n2
= 144 + 24n + n2 = n2 + 24n + 144 9. (2a + b)2 = (2a)2 + 2(2a)(b) + b 2
= 4a2 + 4ab + b2 10. (6x − 3y)2 = (6x)2 − 2(6x)(3y) + (3y)2
= 36x2 − 36xy + 9y2
392
Algebra 1 Worked-Out Solutions
Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 11. (x + 4)2 = x 2 + 2(x)(4) + 42
=
x2
27. 422 = (40 + 2)2
+ 8x + 16
= 402 + 2(40)(2) + 22 = 1600 + 160 + 4
12. (x + 7 + x)2 = (2x + 7)2
=
+ 2(2x)(7) +
(2x)2
= 1764
72
= 4x 2 + 28x + 49
28. 292 = (30 − 1)2
= 302 − 2(30)(1) + 12
13. (7n − 5)2 = (7n)2 − 2(7n)(5) + 52
= 900 − 60 + 1
= 49n2 − 70n + 25 14. (4c + 4d )2 = (4c)2 + 2(4c)(4d ) + (4d )2
= 16c2 + 32cd + 16d 2
= 841 29. 30.52 = (30 + 0.5)2
= 302 + 2(30)(0.5) + 0.52
15. (t − 7)(t + 7) = t 2 − 72
= 900 + 30 + 0.25
= t2 − 49
= 930.25
16. (m + 6)(m − 6) =
m2
−
=
m2
− 36
17. (4x + 1)(4x − 1) =
62
(4x)2
−
1
30. 10 —3
⋅ 9 — = ( 10 + — ) ( 10 − — ) 2 3
= 102 −
12
= 100 −
= 16x2 − 1 18. (2k − 4)(2k + 4) =
= 19. (8 + 3a)(8 − 3a) =
(2k)2 4k2 82
−
(
1
)(
1
2
) () 1
included. (k + 4)2 = k 2 + 2(k)(4) + 42
(3a)2
= k 2 + 8k + 16
9a2
32. There is no middle term in the sum and difference pattern.
(s + 5)(s − 5) = s 2 − 52
= —14 − c2
= s 2 − 25
21. ( p − 10q)( p + 10q) = p2 − (10q)2
= p2 − 100q2 22. (7m + 8n)(7m − 8n) = (7m)2 − (8n)2
23. (−y + 4)(−y − 4) =
49m2
(−y)2
−
−
64n2
42
= y2 − 16 24. (−5g − 2h)(−5g + 2h) = (−5g)2 − (2h)2
= 25g2 − 4h2 25. 16
()
31. The middle term in the square of a binomial pattern was not
− c2
=
1 3
= 99 —89
42
− 16
= 64 − 20. —2 − c —2 + c = —2
−
1 3 1 2 —3 1 —9
⋅ 24 = (20 − 4)(20 + 4) = 202 − 42
33. a. (x + 50)2 = x 2 + 2(x)(50) + 502
= x2 + 100x + 2500 A polynomial that represents the area of the house after the renovation is ( x2 + 100x + 2500 ) square feet. b. x 2 + 100x + 2500 = (15)2 + 100(15) + 2500
= 225 + 1500 + 2500 = 4225 The area of the house after the renovation is 4225 square feet. The original area of the house was 50 2 = 2500 square feet. So, the area of the renovation is 4225 − 2500 = 1725 square feet.
= 400 − 16 = 384 26. 33
⋅ 27 = (30 + 3)(30 − 3) = 302 − 32 = 900 − 9 = 891
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Algebra 1 Worked-Out Solutions
393
Chapter 7 34. a. (100 − x)(100 + x) = 1002 − x 2
37. x 2 − 121 fits the product side of the sum and difference
= 10,000 − x2 The product that represents the area of the new parking lot is ( 10,000 − x2 ) square feet. b. The area of the parking lot decreases. The original area is
10,000 square feet. So, the new area is the original area decreased by x2. c. 10,000 − x 2 = 10,000 − 212
= 10,000 − 441 = 9559 When x = 21, the area of the new parking lot is 9559 square feet. 35. a. The Punnett square shows four possible gene
combinations of the offspring. Of these combinations, one results in albino coloring. So, —14 = 25% of the possible gene combinations result in albino coloring. b. (0.5N +
0.5a)2
=
(0.5N )2
+ 2(0.5N )(0.5a) +
=
0.25N 2
+ 0.5Na + 0.25a2
(0.5a)2
The coefficients show that 25% + 50% = 75% of the possible gene combinations result in normal coloring, while 25% of the possible gene combinations result in albino coloring. 36. a. A = π r2
pattern, so working backward, a and b are the square roots of a2 and b2. —
—
√ x2 = x and √ 121 = 11
So, x2 − 121 = (x + 11)(x − 11). 38. The Punnett square shows four possible gene combinations
of the offspring. Of these combinations, three result in green pods. So, —34 = 75% of the possible gene combinations result in green pods. Alternately, the coefficients of the polynomial model show that GG accounts for 25% of the possible gene combinations, and Gy accounts for 50% of the possible gene combinations. So, 25% + 50% = 75% of the possible gene combinations result in green pods. 39. ( x 2 + 1 )( x 2 − 1 ) = ( x 2 )2 = 12
= x4 − 1 40. ( y3 + 4 )2 = ( y3 )2 + 2( y3 )(4) + 42
= y6 + 8y3 + 16 41. ( 2m2 − 5n2 )2 = ( 2m2 )2 − 2( 2m2 )( 5n2 ) + ( 5n2 )2
= 22m2 ⋅ 2 − 20m2n2 + 52n2 ⋅ 2 = 4m4 − 20m2n2 + 25n4
42. ( r 3 − 6t 4 )( r 3 + 6t 4 ) = ( r 3 )2 − ( 6t 4 )2
= π (6 − x)2
= r 3 ⋅ 2 − 62t 4 ⋅ 2
= π ( 62 − 2(6)(x) + x2 )
= r 6 − 36t 8
= π ( 36 − 12x + x2 ) = π (36) − π (12x) + π ( x2 ) = 36π − 12π x +
π x2
= π x2 − 12π x + 36π A polynomial that represents the area of your pupil is ( π x2 − 12π x + 36π ) square millimeters. b. Let x = 4. Find the area of your pupil before entering
the room.
π x2 − 12π x + 36π = π (4)2 − 12π (4) + 36π = π (16) − 48π + 36π = 16π − 48π + 36π = 4π The area of your pupil before entering the room is 4π millimeters. Let x = 2. Find the area of your pupil after entering the room.
1
(
as 4 +
1 —3
2
not
can be written
) . However, using the square of a binomial pattern ( ) ( ) = 16 + — + —, which is
results in 42 + 2(4) —13 + —13 18 —79 ,
2
( )
43. Your friend is incorrect. The expression 4 —3 2
8 3
1 9
16 —19 .
44. Sample answer: One way to modify the dimensions of the
lake is by increasing one dimension by 4 and decreasing the other dimension by 4, which can be modeled by (x + 4)(x − 4). This follows a sum and difference pattern. A second way to modify the dimensions of the lake is by increasing both dimensions by 2, which can be modeled by (x + 2)2. This follows a square of a binomial pattern. A third way to modify the dimensions of the lake is by decreasing both dimensions by 2, which can be modeled by (x − 2)2. This also follows a square of a binomial pattern, but this one has subtraction.
π x2 − 12π x + 36π = π (2)2 − 12π (2) + 36π = π (4) − 24π + 36π = 4π − 24π + 36π = 16π The area of your pupil after entering the room is 16π 16π millimeters. So, the area of your pupil is — = 4 times 4π greater after entering the room than before.
394
Algebra 1 Worked-Out Solutions
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Chapter 7 45. Let 9x 2 − 48x + k = a2 − 2ab + b 2.
So,
50. 49s + 35t = 7(7s + 5t)
9x2 = a2
—
51. 15x − 10y = 5(3x − 2y)
—
√9x2 = √a2 3x = a. Then,
Finally,
7.4 Explorations (p. 377)
48x = 2ab 48x = 2(3x)b 48x = 6xb 48x 6xb —=— 6x 6x 8 = b.
(x − 1)(x − 3) = 0 ← Factored Form
1. a. C; 5;
F
O
I
L
x(x) + x(−3) + (−1)(x) + (−1)(−3) = 0 x2 − 3x − x + 3 = 0 x2 − 4x + 3 = 0 ← Standard Form, C
k = b2 k = 82 k = 64.
−3
So, when k = 64, 9x2 − 48x + k = 9x2 − 48x + 64 is the square of the binomial 3x − 8. 46. (x + 1)3 = (x + 1)(x + 1)(x + 1)
= [x(x + 1) + 1(x + 1)](x + 1) = [x(x) + x(1) + 1(x) + 1(1)](x + 1) = ( x2 + x + x + 1 )(x + 1) = ( x2 + 2x + 1 )(x + 1) = x2(x + 1) + 2x(x + 1) + 1(x + 1) = x2(x) + x2(1) + 2x(x) + 2x(1) + 1(x) + 1(1) = x3 + ( x2 + 2x2 ) + (2x + x) + 1 = x3 + 3x2 + 3x + 1 (x + 2)3 = (x + 2)(x + 2)(x + 2) = [x(x + 2) + 2(x + 2)](x + 2) = [x(x) + x(2) + 2(x) + 2(2)](x + 2) = ( x2 + 2x + 2x + 4 )(x + 2) = ( x2 + 4x + 4 )(x + 2) = x2(x + 2) + 4x(x + 2) + 4(x + 2) = x2(x) + x2(2) + 4x(x) + 4x(2) + 4(x) + 4(2) = x3 + 2x2 + 4x2 + 8x + 4x + 8 = x3 + 6x2 + 12x + 8 So, (a + b)3 = a3 + 3a2b + 3ab2 + b3. 47. Sample answer: The statement (a + b)(a − b) < (a − b)2 <
(a + b)2 is true when a = 3 and b = 4. Check (a + b)(a − b) < (a − b)2 ? (3 + 4)(3 − 4) < (3 − 4)2 ? (7)(−1) < (−1)2 −7 <
1
< (a + b)2 ? < (3 + 4)2 ? < (7)2 <
−3
x2 − 4x = −3 ← Nonstandard Form, 5 So, the equation (x − 1)(x − 3) = 0 is equivalent to x2 − 4x + 3 = 0 (C) and x2 − 4x = −3 (5). (x − 2)(x − 3) = 0 ← Factored Form
b. D; 1;
F
O
I
L
x(x) + x(−3) + (−2)(x) + (−2)(−3) = 0 x2 − 3x − 2x + 6 = 0 x2 − 5x + 6 = 0 ← Standard Form, D −6
−6
x2 − 5x = −6 ← Nonstandard Form, 1 So, the equation (x − 2)(x − 3) = 0 is equivalent to x2 − 5x + 6 = 0 (D) and x2 − 5x = −6 (1). (x + 1)(x − 2) = 0 ← Factored Form
c. A; 3;
F
O
I
L
x(x) + x(−2) + 1(x) + 1(−2) = 0 x2 − 2x + x − 2 = 0 x2 − x − 2 = 0 ← Standard Form, A +2
+2
x2 − x = 2 ← Nonstandard Form, 3 So, the equation (x + 1)(x − 2) = 0 is equivalent to x2 − x − 2 = 0 (A) and x2 − x = 2 (3).
49 ✓
Maintaining Mathematical Proficiency 48. 12y − 18 = 6(2y − 3) 49. 9r + 27 = 9(r + 3)
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Algebra 1 Worked-Out Solutions
395
Chapter 7 (x − 1)(x + 2) = 0 ← Factored Form
d. B; 4;
F
O
I
L
x(x) + x(2) + (−1)(x) + (−1)(2) = 0
d.
x2 + 2x − x − 2 = 0 x2 + x − 2 = 0 ← Standard Form, B x(x + 1) = 2 ← Nonstandard Form, 4 x(x) + x(1) = 2 x2
+x=2
e.
x2 + x − 2 = 2 − 2
So, the equation (x − 1)(x + 2) = 0 is equivalent to x2 + x − 2 = 0 (B) and x(x + 1) = 2 (4). (x + 1)(x − 3) = 0 ← Factored Form
F
O
I
L
x(x) + x(−3) + 1(x) + 1(−3) = 0 x2 − 2x − 3 = 0 ← Standard Form, E (x − 1)2 = 4 ← Nonstandard Form, 2 x2 − 2(x)(1) + 12 = 4
So, the equation (x + 1)(x − 3) = 0 is equivalent to x2 − 2x = 3 (E) and (x − 1)2 = 4 (2).
b.
c.
396
(−3)(−4) ≠ 0 False
(−2)(−3) ≠ 0 False
3
4
5
6
(−1)(−2) ≠ 0 False
(0)(−1) = 0 True
(1)(0) = 0 True
(2)(1) ≠ 0 False
x=
1
2
(x − 5)(x − 6) = 0
(−4)(−5) ≠ 0 False
(−3)(−4) ≠ 0 False
3
4
5
6
f.
x=
1
2
(x − 6)(x − 1) = 0
(−5)(0) = 0 True
(−4)(1) ≠ 0 False
3
4
5
6 (0)(5) = 0 True
If the product of two values is 0, then at least one of the values must be 0. If (x − a) is a factor of an equation, then x = a is a solution. not change the value, but adding 1 increases the value.
−4
x2 − 2x − 3 = 0 ← Standard Form, E
a.
(x − 4)(x − 5) = 0
3. a. When you add 0 to a number n, you get n. Adding 0 does
x2 − 2x + 1 = 4
2.
2
(−3)(2) ≠ 0 (−2)(3) ≠ 0 (−1)(4) ≠ 0 False False False
x2 − 3x + x − 3 = 0
−4
1
(−2)(−3) ≠ 0 (−1)(−2) ≠ 0 (0)(−1) = 0 (1)(0) = 0 False False True True
x2 + x − 2 = 0 ← Standard Form, B
e. E; 2;
x=
b. If the product of two numbers is 0, then at least one of the
numbers is 0. The product of 0 and any number is always 0, never 1. c. The square of 0 is equal to itself, and the square of 1 is
equal to itself. Both are true: 02 = 0 and 12 = 1.
x=
1
2
(x − 1)(x − 2) = 0
(0)(−1) = 0 True
(1)(0) = 0 True
d. When you multiply a number n by 1, you get n. The
3
4
5
6
(2)(1) ≠ 0 False
(3)(2) ≠ 0 False
(4)(3) ≠ 0 False
(5)(4) ≠ 0 False
x=
1
2
(x − 2)(x − 3) = 0
(−1)(−2) ≠ 0 False
(0)(−1) = 0 True
3
4
5
6
(1)(0) = 0 True
(2)(1) ≠ 0 False
(3)(2) ≠ 0 False
(4)(3) ≠ 0 False
x=
1
2
(x − 3)(x − 4) = 0
(−2)(−3) ≠ 0 False
(−1)(−2) ≠ 0 False
3
4
5
6
(0)(−1) = 0 True
(1)(0) = 0 True
(2)(1) ≠ 0 False
(3)(2) ≠ 0 False
Algebra 1 Worked-Out Solutions
product of any number and 1 is the number. On the other hand, the product of any number and 0 is 0. e. When you multiply a number n by 0, you get 0. The
product of any number and 0 is 0. On the other hand, the product of any number and 1 is the number. f. The opposite of 0 is equal to itself. Zero is neither positive
nor negative. So, it is its own opposite. The opposite of 1, however, is −1. 4. When an equation is in factored form, the product of the
factors equals 0. So, set each polynomial factor equal to 0, and solve. 5. b; If the product of two numbers is 0, then at least one of
the numbers is 0. This property is called the Zero-Product Property because it describes what happens when you have a product that is equal to zero. The Zero-Product Property is used in algebra to solve polynomial equations by setting each polynomial factor equal to 0. It is important because it provides an easy way to solve polynomial equations.
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Chapter 7 7.4 Monitoring Progress (pp. 378–380)
4. (3s + 5)(5s + 8) = 0
1. x(x − 1) = 0
x=0
or
3s + 5 = x−1= 0
−5
+1 +1 x(x − 1) = 0 ? 1(1 − 1) = 0 ? 1(0) = 0
0=0✓
or
t+2= −2
? 5 5 −— + 8 = 0 3 ? 25 (−5 + 5) −— + 8 = 0 3 1 ? 0 −— = 0 3 0=0✓
(
0 −2
[ ( ) ][ ( ) ]
? 8 8 3 −— + 5 5 −— + 8 = 0 5 5 ? 24 −— + 5 (−8 + 8) = 0 5 ? 1 — (0) = 0 5 0=0✓
3t(t + 2) = 0 ? 3(−2)(−2 + 2) = 0 ? −6(0) = 0
0=0✓
(
8 5 The roots are s = −— and s = −—. 3 5
3. (z − 4)(z − 6) = 0
or
z−6= 0
+4 +4
+6 +6
z= 4
z= 6
Check (z − 4)(z − 6) = 0 ? (4 − 4)(4 − 6) = 0 ? 0(−2) = 0 0=0✓ The roots are z = 4 and z = 6.
)
()
0=0✓
The roots are t = 0 and t = −2. z−4= 0
) ( )
(3s + 5)(5s + 8) = 0
t=−2
Check 3t(t − 2) = 0 ? 3(0)(0 + 2) = 0 ? 0(2) = 0
−8
5 3 −— + 5 3
2. 3t(t + 2) = 0
3t = 0
−8
[ ( ) ][ ( ) ]
0=0✓
The roots are x = 0 and x = 1.
3t 0 —=— 3 3 t=0
−5
0
3s = − 5 5s = − 8 5s −8 3s −5 —=— —=— 3 5 5 3 8 5 s = −— s = −— 3 5 Check (3s + 5)(5s + 8) = 0
x= 1 Check x(x − 1) = 0 ? 0(0 − 1) = 0 ? 0(−1) = 0
5s + 8 =
or
0
(b + 7)2 = 0
5.
(b + 7)(b + 7) = 0
(z − 4)(z − 6) = 0 ? (6 − 4)(6 − 6) = 0 ? 2(0) = 0 0=0✓
b+7= −7
0
b+7=
or
−7
−7
b = −7
0 −7
b = −7
Check (b + 7)2 = 0 ? (−7 + 7)2 = 0 ? (0)2 = 0 0=0✓ The equation has repeated roots of b = −7.
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Algebra 1 Worked-Out Solutions
397
Chapter 7 6. (d − 2)(d + 6)(d + 8) = 0
d−2= 0
d+6=
or
+2 +2
−6
d=2
0 or −6
d+8= −8
d = −6
0 −8
d = −8
Check (d − 2)(d + 6)(d + 8) = 0 ? (2 − 2)(2 + 6)(2 + 8) = 0 ? 0(8)(10) = 0
0=0✓ 7. The GCF of 8 and −24 is 8. The GCF of y2 and y is y. So,
the greatest common monomial factor of the terms is 8y. So, 8y2 − 24y = 8y( y − 3).
2x = 1
4x2 = 2x 1 2 ? 1 4 — =2 — 2 2 1 ? 4 — =1 0=0✓ 4 1=1✓ 1 The roots are x = 0 and x = —. 2
0
()
1
y = −—2 (x + 4)(x − 4)
⋅ ( −— )(x + 4)(x − 4) 1 2
0 = (x + 4)(x − 4)
−5
x+4=
a = −5 Check a2 + 5a = 0 ? (0)2 + 5(0) = 0 ? 0+0=0
() ()
11. Let y = 0.
−2(0) = −2 −5
+1
1
+ 5a = 0
a2 + 5a = 0 ? (−5)2 + 5(−5) = 0 ? 25 − 25 = 0
0=0✓
0=0✓
The roots are a = 0 and a = −5.
−4
0
or
x−4= 0
−4
+4 +4
x = −4
x= 4
So, the width of the entrance at ground level is ∣ − 4 − 4 ∣ = 8 feet. 7.4 Exercises (pp. 381–382) Vocabulary and Core Concept Check
9. 3s2 − 9s = 0
1. The product of 3x and x − 6 equals 0. According to the
3s(s − 3) = 0 or
+1
0
0 = −—2 (x + 4)(x − 4)
a(a + 5) = 0 a+5=
2x − 1 =
Check 4x2 = 2x ? 4(0)2 = 2(0) ? 4(0) = 0
The roots are d = 2, d = −6, and d = −8.
3s = 0 3s 0 —=— 3 3 s=0
or
1 2 1 x=— 2
(d − 2)(d + 6)(d + 8) = 0 ? (−8 − 2)(−8 + 6)(−8 + 8) = 0 ? (−10)(−2)(0) = 0
or
2x(2x − 1) = 0
2x 2
0=0✓
a=0
4x2 − 2x = 0
—=—
(d − 2)(d + 6)(d + 8) = 0 ? (−6 − 2)(−6 + 6)(−6 + 8) = 0 ? (−8)(0)(2) = 0
8.
4x2 − 2x = 2x − 2x
2x = 0 2x 0 —=— 2 2 x=0
0=0✓
a2
4x2 = 2x
10.
s−3= +3
Zero-Product Property, at least one of the factors equals 0. So, let 3x = 0 and x − 6 = 0. Then solve each equation to get the solutions x = 0 and x = 6.
0 +3
s=3
Check 3s2 − 9s = 0 ? 3(0)2 − 9(0) = 0 ? 0−0=0 0=0✓
3s2 − 9s = 0 ? 3(3)2 − 9(3) = 0 ? 3(9) − 27 = 0 ? 27 − 27 = 0 0=0✓
The roots are s = 0 and s = 3.
398
Algebra 1 Worked-Out Solutions
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Chapter 7 2. The one that is different is “Find the value of k for which
(2k + 4) + (k − 3) = 0.”
6. −2v(v + 1) = 0
−2v = 0 or −2v 0 —=— −2 −2 v=0
(2k + 4) + (k − 3) = 0 (2k + k) + (4 − 3) = 0 3k + 1 = −1
0
v+1= −1
v = −1
−1
7. (s − 9) (s − 1) = 0
(2k + 4)(k − 3) = 0
8. ( y + 2) ( y − 6) = 0
−4
s−9= +9
−4
−2
k=3
0 +6
y=6
9. (2a − 6) (3a + 15) = 0
2a − 6 = 0
x+7= −7
−7
2a = 6
3a = −15
2a 6 2 2 a=3
—=—
3a −15 3 3 a = −5
10. (4q + 3) (q + 2) = 0
4q + 3 =
4. r(r − 10) = 0
−3
0
r = 10
q+2=
or
−3
−2
−3 4 3 q = −— 4
3 The roots are q = −— and q = −2. 4
0
t=5
The roots are t = 0 and t = 5.
−2
q = −2
4q 4
5. 12t(t − 5) = 0
+5
0
—=—
The roots are r = 0 and r = 10. t−5=
0
4q = −3
+ 10 + 10
+5
0 − 15
The roots are a = 3 and a = −5.
x = −7 The roots are x = 0 and x = −7.
or
− 15
—=—
0
r − 10 =
3a + 15 =
or
+6 +6
3. x(x + 7) = 0
12t = 0 0 12t —=— 12 12 t=0
+6
y = −2
Monitoring Progress and Modeling with Mathematics
or
y−6=
or
0 −2
The roots are y = −2 and y = 6.
The roots are k = −2 and k = 3.
r=0
0 +1
s=1
y+2=
+3 +3
2k −4 —=— 2 2 k = −2
or
+1
s=9
2k = −4
x=0
s−1=
or
0 +9
The roots are s = 9 and s = 1.
k−3= 0
or
0
−1
The roots are v = 0 and v = −1.
3k = −1 3k −1 —=— 3 3 1 k = −— 3 1 The solution is k = −—. 3 2k + 4 =
0
(5m + 4)2 = 0
11.
(5m + 4) (5m + 4) = 0 5m + 4 = −4
0 −4
5m = −4
or
5m + 4 = −4
0 −4
5m = −4
5m −4 —=— 5 5 4 m = −— 5
5m −4 —=— 5 5 4 m = −— 5 4 The equation has repeated roots of m = −—. 5
Copyright © Big Ideas Learning, LLC All rights reserved.
Algebra 1 Worked-Out Solutions
399
Chapter 7 (h − 8)2 = 0
12.
17.
(h − 8) (h − 8) = 0 h−8=
h−8=
or
0
+8 +8
+8
h=8
r−4=
0 +8
+4
h=8
0
7−g=
or
+ 2g
+g
3 = 2g
+4
0
w=0
or w − 6 = +6
15 − 5c = 2 + 4d =
−2
−2
−4d = −2
− 15
0 −2
4d = −2
−2
5p 5
0 5
—=—
p=0
+3
0
0
z=1
or
p+7= 0
2p = 3 2p 3 —=— 2 2 3 p=— 2
3 The roots are p = 0, p = —, and p = −7. 2
400
—=—
−5 5
—=—
Algebra 1 Worked-Out Solutions
5c 5
−c −1
c = −1
)
2 3
or
0 −2
−6 −1
c=6
−7 −7 p = −7
or
2 3 3 2 3 — —n = — ( −6 ) 2 3 2
−2 −1
−n −1
2 6 + —n = 0 3 −6 −6 —n = −6
−n = −2
⋅
—=—
n=2
+3
0 −6
−15 −5
2−n=
16. 5p(2p − 3)(p + 7) = 0
2p − 3 =
−6
−c = −6
(
The roots are z = 0, z = −2, and z = 1.
or
or −c + 6 =
20. (2 − n) 6 + —n (n − 2)= 0
+1 +1
z = −2
0 −5
5c = −5
c=3
z−1=
−2
−5
−5c = −15
−2 or
5c + 5 =
The roots are c = 3, c = −1, and c = 6.
15. z (z + 2)(z − 1) = 0
0
or
0 − 15
−5c −5
4d −2 —=— 4 4 1 d = −— 2 1 1 The roots are d = — and d = −—. 2 2
z+2=
0 +6
w=6
—=—
−4d −2 —=— −4 −4 1 d=— 2
5p = 0
+6
19. (15 − 5c)(5c + 5)(−c + 6) = 0
14. (2 − 4d ) (2 + 4d ) = 0
or
w−6=
or
0 +6
One root of the equation is w = 0. The equation also has repeated roots of w = 6.
3 The roots are g = — and g = 7. 2
or
−8
r = −8
w=6
0
0
−8
w(w − 6)(w − 6) = 0
7=g
3 2
z=0
r+8=
18. w(w − 6)2 = 0
+g
—=g
−2
or
r=4
3 2g —=— 2 2
2 − 4d =
0 +4
One root of the equation is r = −8. The equation also has repeated roots of r = 4.
13. (3 − 2g) (7 − g) = 0
3 − 2g =
r−4=
or
0 +4
r=4
The equation has repeated roots of h = 8.
+ 2g
(r − 4)2(r + 8) = 0 (r − 4)(r − 4)(r + 8) = 0
n−2= 0 +2 +2 n=2
⋅
n = −9
One root of the equation is n = −9. The equation also has repeated roots of n = 2. 21. y = (x − 8)(x + 8)
0 = (x − 8)(x + 8) x−8= +8
0 +8
x=8
or
x+8= −8
0 −8
x = −8
The x-coordinates of the points where the graph crosses the x-axis are the roots x = 8 and x = −8.
Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 22. y = ( x + 1 )( x + 7 )
32. 6m2 + 12m = 0
0 = ( x + 1 )( x + 7 )
6m( m + 2 ) = 0
x+1=0 −1
x+7=0
or
−1
−7
x = −1
6m = 0 6m 0 —=— 6 6 m=0
−7
x = −7
The x-coordinates of the points where the graph crosses the x-axis are the roots x = −1 and x = −7.
+14
5c = 0 5c 0 —=— 5 5 c=0
+5
x = 14
+5
x=5
The x-coordinates of the points where the graph crosses the x-axis are the roots x = 14 and x = 5. 24.
y = −0.2( x + 22 )( x − 15 ) 0 = −0.2( x + 22 )( x − 15 ) −0.2( x + 22 )( x − 15 ) 0 — = —— −0.2 −0.2 0 = ( x + 22 )( x − 15 ) x + 22 = 0 −22
or
+15
x = −22
2q 0 2 2 q=0
9=q
3n2
− 9n = 9n − 9n
3n2
− 9n = 0
3n( n − 3 ) = 0 3n = 0
or
n−3=0
3n 0 —=— 3 3 n=0
+3 +3 n=3
The roots are n = 0 and n = 3. 36.
−28r = 4r2 −28r + 28r = 4r2 + 28r 0 = 4r2 + 28r 0 = 4r( r + 7 )
31. 4p2 − p = 0
p( 4p − 1 ) = 0 or
+q +q
3n2 = 9n
35.
28. 20x3 + 30x2 = 10x2( 2x + 3 )
p=0
9−q=0
The roots are q = 0 and q = 9.
27. 3y3 − 9y2 = 3y2( y − 3 )
30. 12a4 + 8a = 4a( 3a3 + 2 )
or
—=—
26. 6d2 − 21d = 3d( 2d − 7 )
29. 5n6 + 2n5 = n5( 5n + 2 )
−5
2q( 9 − q ) = 0
The x-coordinates of the points where the graph crosses the x-axis are the roots x = −22 and x = 15. 25. 5z2 + 45z = 5z( z + 9 )
−5
2q = 0
+15
x = 15
5 + 2c = 0
34. 18q − 2q2 = 0
x − 15 = 0
−22
or
2c = −5 2c −5 —=— 2 2 −5 c=— 2 5 The roots are c = 0 and c = −—. 2
x−5=0
+14
m = −2
5c( 5 + 2c ) = 0
0 = ( x − 14 )( x − 5 ) or
−2 −2
33. 25c + 10c2 = 0
0 = −( x − 14 )( x − 5 ) ( −1 0 ) = −1 ⋅ [ −( x − 14 )( x − 5 ) ] x − 14 = 0
m+2=0
The roots are m = 0 and m = −2.
y = −( x − 14 )( x − 5 )
23.
or
4p − 1 = 0 +1 +1 4p = 1 4p 1 —=— 4 4 1 p=— 4
4r = 0 4r 0 —=— 4 4 r=0
or
r+7=0 −7
−7
r = −7
The roots are r = 0 and r = −7.
1 The roots are p = 0 and p = —. 4
Copyright © Big Ideas Learning, LLC All rights reserved.
Algebra 1 Worked-Out Solutions
401
Chapter 7 37. The other factor, 6x, should also be set equal to 0. Then
each equation should be solved to get two roots of the original equation. 6x( x + 5 ) = 0 x=0
x+5=0
or
6x 0 —=— 6 6 x=0
−5 −5 x = −5
b. Let x = 0.
2 y = −—( x + 315 )( x − 315 ) 315 2 y = −—( 0 + 315 )( 0 − 315 ) 315 2 y = −—( 315 )( −315 ) 315 = ( −2 )( −315 ) = 630
The roots are x = 0 and x = −5.
So, the height of the arch is 630 feet.
38. Each side of the equation cannot be divided by y because
y could be 0, and division by 0 is undefined. Instead, 21y should be subtracted from each side, so that the equation is 3y2 − 21y = 0, and the greatest monomial factor should be factored out on the left side of the equal sign. Then each factor should be set equal to zero, and each equation should be solved to find both roots of the original equation.
41. y = −16x2 + 4.8x
0 = −16x2 + 4.8x 0 = x( −16x + 4.8 ) x=0
or
−4.8
3y2 − 21y = 21y − 21y 3y2 − 21y = 0 3y( y − 7 ) = 0
Because y = 0, the roots are the x-values that represent the times when the penguin is at water level. So, when x = 0, the penguin emerges from the water. Then it is above the water until 0.3 second later, the moment it breaks the surface of the water again after the leap. So, the penguin is airborne for 0.3 second with each leap.
y−7=0
or
3y 0 3 3 y=0
—=—
+7 +7 y=7
The roots are y = 0 and y = 7.
42. The point of intersection on the positive x-axis is farther
39. Let y = 0.
from the origin than the point of intersection on the negative x-axis. So, the positive solution to the equation is 5, because it is the one with the greater absolute value. Therefore, the equation is y = ( x − 5 )( x + 3 ).
11 y = −—( x − 4 )( x − 24 ) 50 11 0 = −—( x − 4 )( x − 24 ) 50 50( ) 50 11 ( −— 0 = −— −— x − 4 )( x − 24 ) 11 11 50 0 = ( x − 4 )( x − 24 )
( )
x−4=0 +4
−4.8
−16x = −4.8 −16x −4.8 — = — −16 −16 x = 0.3
3y2 = 21y
3y = 0
−16x + 4.8 = 0
43. The graph has two x-intercepts because x-intercepts occur
when y = 0, and this equation has two roots (one of which is a repeated root) when y = 0.
x − 24 = 0
or
+4
+24
x=4
44. no; The equation y = ( x − a )( x − b ) will not have two
+24
x-intercepts if a = b. In this case, there is a repeated root, and the graph will only have one x-intercept.
x = 24
So, the width of the tunnel at ground level is ∣ 24 − 4 ∣ = 20 feet. 40. a. Let y = 0.
2 y = −—( x + 315 )( x − 315 ) 315 2 0 = −—( x + 315 )( x − 315 ) 315 315 315 2 −—( 0 ) = −— −— ( x + 315 )( x − 315 ) 2 2 315 0 = ( x + 315 )( x − 315 )
)
(
x + 315 = 0 −315
or
x − 315 = 0
−315
+315 +315
x = −315
x = 315
45. no; The equation ( x2 + 3 )( x4 + 1 ) = 0 does not have any
real roots. Roots will occur if x3 + 3 = 0 or x4 + 1 = 0. However, solving these equations results in x2 = −3 or x4 = −1, and even powers of any number cannot be negative. x3 + 3 = 0
x4 + 1= 0
−3
−3
−1
−1
x3
= −3
x4
= −1
46. Sample answer: A polynomial equation of degree 4 that only
has three roots must have a repeated root. So, one possible equation is ( x − 1 )( x − 2 )2( x − 3 ) = 0.
So, the width of the arch at ground level is ∣ −315 − 315 ∣ = 630 feet.
402
Algebra 1 Worked-Out Solutions
Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 52. 48
47. a. ( x + y )( 2x − y ) = 0
x+y=0 x+y−y=0−y
1( 48 ) = 48 2( 24 ) = 48
2x − y = 0
or
2x − y + y = 0 + y
x = −y
2x = y
y 2 1 x = —y 2 1 So, the roots of the equation occur when x = −y and x = —y. 2 2x 2
—=—
b. ( x2 − y2 )( 4x + 16y ) = 0
x2 − y2 = 0
4x + 16y = 0
or
x2 − y2 + y2 = 0 + y2
4x − 4x + 16y = 0 − 4x
x2 = y2 —
16y = −4x 16y −4x −4 −4 −4y = x
—
± √ x2 = ± √ y2
—= —
x=±y
So, the roots of the equation occur when x = ± y and x = −4y. 48.
( 3x
− 81 ) = 0
4x−5 − 16 = 0
or
( 4x−5
− 16 )
+16 +16
3x − 81 = 0 +81
+81
4x−5 = 16
3x = 81
4x−5 = 42
3 x = 34
x−5=2
x=4
+5 +5 x=7 The solutions are x = 7 and x = 4.
3( 16 ) = 48 4( 12 ) = 48 6( 8 ) = 48 So, the factor pairs of 48 are 1 and 48, 2 and 24, 3 and 16, 4 and 12, and 6 and 8. 7.1–7.4 What Did You Learn? (p. 383) 1. Sample answer: Use x + 4 to represent each side of the
square. So, an expression that represents the area is ( x + 4 )2. Then use the square of a binomial pattern to simplify this expression into a polynomial. As another method, you could calculate the area of each shaded region and add those areas together. 2. Sample answer: The solutions are the constant terms with
the opposite sign. This method does not work when the coefficient is not 1. 7.1–7.4 Quiz (p. 384) 1. The polynomial −8q3 is in standard form.
The only term has a degree of 3. So, the degree of the polynomial is 3. The leading coefficient is −8. The polynomial has 1 term. So, it is a monomial. 2. You can write the polynomial 9 + d 2 − 3d in standard form
as d 2 − 3d + 9.
The greatest degree is 2. So, the degree of the polynomial is 2. Maintaining Mathematical Proficiency 49. 10
1( 10 ) = 10 2( 5 ) = 10 So, the factor pairs of 10 are 1 and 10, and 2 and 5. 50. 18
1( 18 ) = 18 2( 9 ) = 18 3( 6 ) = 18 So, the factor pairs of 18 are 1 and 18, 2 and 9, and 3 and 16. 51. 30
1( 30 ) = 30 2( 15 ) = 30 3( 10 ) = 30
The leading coefficient is 1. The polynomial has 3 terms. So, it is a trinomial. 2 3
5 m6 + 2 m4. as − — — 6 3 The greatest degree is 6. So, the degree of the polynomial is 6. 5 The leading coefficient is −—. 6 The polynomial has 2 terms. So, it is a binomial. 4. You can write the polynomial −1.3z + 3z4 + 7.4z2 in
standard form as 3z4 + 7.4z2 − 1.3z.
The greatest degree is 4. So, the degree of the polynomial is 4. The leading coefficient is 3. The polynomial has 3 terms. So, it is a trinomial. 5. ( 2x2 + 5 ) + ( −x2 + 4 ) = 2x2 − x2 + 5 + 4
= ( 2x2 − x2 ) + ( 5 + 4 )
5( 6 ) = 30 So, the factor pairs of 30 are 1 and 30, 2 and 15, 3 and 10, and 5 and 6.
Copyright © Big Ideas Learning, LLC All rights reserved.
5 6
3. You can write the polynomial — m4 − — m6 in standard form
= x2 + 9 The sum is
x2
+ 9.
Algebra 1 Worked-Out Solutions
403
Chapter 7 6. ( −3n2 + n ) − ( 2c2 − 7 ) = −3n2 + n − 2n2 + 7
The difference is −p2 + 4p
−5n2
5x(x − 3) = 0
= −5n2 + n + 7
5x = 0 5x 0 —=— 5 5 x=0
+ n + 7.
)−( ( = −p2 + 4p − p2 + 3p − 15
7.
p2 − 3p + 15
)
3=0
8−g=0
The difference is −2p2 + 7p − 15. 8. ( a2 − 3ab + b2 ) + ( −a2 + ab + b2 )
= a2 − a2 − 3ab + ab + b2 + b2 = ( a2 − a2 ) + ( −3ab + ab ) + ( b2 + b2 ) = 0 − 2ab + 2b2
+g +g
+g +g
8=g
8=g
The equation has repeated roots of g = 8. 17. (3p + 7)(3p − 7)(p + 8) = 0
−7
The sum is −2ab + 2b2. 9. ( w + 6 )( w + 7 ) = w( w + 7 ) + 6( w + 7 )
= w( w ) + w( 7 ) + 6( w ) + 6( 7 ) = w2 + 7w + 6w + 42 = w2 + 13w + 42 + 13w + 42.
8−g=0
or
3p + 7 =
= −2ab + 2b2
0
or
3p − 7 =
−7
+7
or
0 +7
O
I
L
= 3( 2d ) + 3( −5 ) + ( −4d )( 2d ) + ( −4d )( −5 ) = 6d − 15 − 8d 2 + 20d = −8d 2 + ( 6d + 20d ) − 15
−8
3p = −7 7 3p −— = −— 3 3 7 p = −— 3
3p = 7 3p 7 —=— 3 3 7 p=— 3 7 7 The roots are p = −—, p = —, and p = −8. 3 3 −3y = 0 or y − 8 = 0 or 0 −3y —=— +8 +8 −3 −3 y=0 y=8
−1
The product is −8d 2 + 26d − 15. y2 + 2y − 3 y+9
×
—— 2 + 18y − 27 9y y3 + 2y2 − 3y
—— 3 + 11y2 + 15y − 27 y—— The product is y3 + 11y2 + 15y − 27. 12. (3z − 5)(3z + 5) = (3z)2 − 52
= 9z2 − 25 The product is 9z2 − 25. 13. (t + 5)2 = t 2 + 2(t)(5) + 52
=t 2
+ 10t + 25
The product is t 2 + 10t + 25. 14. (2q − 6)2 = (2q)2 − 2(2q)(6) + 62
= 4q2 − 24q + 36 The product is 4q2 − 24q + 36.
404
0 −8
p = −8
2y + 1 = 0
= −8d 2 + 26d − 15
11.
p+8=
18. −3y(y − 8)(2y + 1) = 0
10. ( 3 − 4d )( 2d − 5 ) F
+3 +3
16. (8 − g)(8 − g) = 0
= −2p2 + 7p − 15
The product is
x−3=0
or
The roots are x = 0 and x = 3.
= ( −p2 − p2 ) + ( 4p + 3p ) − 15
w2
15. 5x2 − 15x = 0
= ( −3n2 − 2n2 ) + n + 7
Algebra 1 Worked-Out Solutions
1 The roots are y = 0, y = 8, and y = −—. 2
−1
2y = −1 2y −1 —=— 2 2 1 y = −— 2
19. a. P = 2ℓ + 2w
= 2(x + 72 + x) + 2(x + 48 + x) = 2(2x + 72) + 2(2x + 48) = 2(2x) + 2(72) + 2(2x) + 2(48) = 4x + 144 + 4x + 96 = 8x + 240 A polynomial that represents the perimeter of the blanket including the fringe is (8x + 240) inches. b. A = ℓw
= (x + 72 + x)(x + 48 + x) = (2x + 72)(2x + 48) F
O
I
L
= 2x(2x) + 2x(48) + 72(2x) + 72(48) = 4x2 + 96x + 144x + 3456 = 4x2 + 240x + 3456 A polynomial that represents the area of the blanket including the fringe is (4x2 + 240x + 3456) square inches. Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 c. P = 8x + 240
Section 7.5
= 8(4) + 240
7.5 Explorations (p. 385)
= 32 + 240
1. a.
= 272 A = 4x2 + 240x + 3456 = 4(4)2 + 240(4) + 3456 = 4(16) + 960 + 3456
Area = x 2 − 3x + 2 = (x − 1)(x − 2)
= 64 + 960 + 3456 b.
= 4480 When the width of the fringe is 4 inches, the perimeter of the blanket is 272 inches and the area is 4480 square inches. 20. a. 1000(1 + r)2 = 1000(1 + r)(1 + r)
F
O
I
Area = x 2 + 5x + 4 = (x + 1)(x + 4)
L
= 1000[ 1(1) + 1(r) + r(1) + r(r) ]
c.
= 1000(1 + r + r + r2) = 1000(r2 + 2r + 1) = 1000(r2) + 1000(2r) + 1000(1) = 1000r2 + 2000r + 1000 The polynomial in standard form that represents the balance of your bank account after 2 years is (1000r2 + 2000r + 1000) dollars. b.
1000r2 + 2000r + 1000 = 1000(0.03)2 + 2000(0.03) + 1000 = 1000(0.0009) + 60 + 1000
Area = x 2 − 7x + 12 = (x − 4)(x − 3) d.
= 0.9 + 60 + 1000 = 1060.9 When the interest rate is 3%, the balance after 2 years is $1060.90. c. 1000(1 + r3) = 1000(1 + 0.03)3
= 1000(1.03)3
Area = x 2 + 7x + 12 = (x + 4)(x + 3)
= 1000(1.092727)
2. Arrange algebra tiles that model the trinomial into a
= 1092.73 The balance after 3 years is only $1092.73. So, you do not have enough money for the $1100 guitar.
rectangular array, use additional algebra tiles to model the dimensions of the rectangle, then write the polynomial in factored form using the dimensions of the rectangle.
5
y = −— 216 (x − 72)(x + 72)
21.
0= 216
5 −— 216 (x 216
(
3. Find two integer factors of c that have a sum of b, then write
the binomial factors by adding each integer factor to x.
− 72)(x + 72) 5
)
7.5 Monitoring Progress (pp. 386–388)
— — −— 50 (0) = − 5 − 216 (x − 72)(x + 72)
1. x 2 + 7x + 6
0 = (x − 72)(x + 72) x − 72 = + 72
0 + 72
or
x + 72 = − 72
x = 72
0 − 72
x = −72
So, the width of the bunker at ground level is ∣ −72 − 72 ∣ = 144 inches.
Copyright © Big Ideas Learning, LLC All rights reserved.
Factors of 6
Sum of factors
1, 6
7
2, 3
5
So, x 2 + 7x + 6 = (x + 1)(x + 6).
Algebra 1 Worked-Out Solutions
405
Chapter 7 2. x 2 + 9x + 8
9.
200 = (s − 30)(s − 40) 200 = s(s) + s(−40) + (−30)(s) + (−30)(−40)
Factors of 8
Sum of factors
1, 8
9
200 = s 2 − 40s − 30s + 1200
2, 4
6
200 = s 2 − 70s + 1200 − 200
So, x 2 + 9x + 8 = (x + 1)(x + 8). 3. w 2 − 4w + 3
0 = (s − 50)(s − 20)
Factors of 3
−1, −3
s − 50 =
−4
+ 50
Sum of factors
−1, −35
−5, −7
−36
−12
Sum of factors So,
−1, −24 −2, −12 −3, −8
Sum of factors
−25
−14
−4, −6
−11
−10
So, x 2 − 14x + 24 = (x − 2)(x − 12).
Factors of −15
−1, 15
1, −15
−3, 5
3, −5
Sum of factors
14
−14
2
−2
+ 2x − 15 = (x − 3)(x + 5).
that are opposite signs. When factoring x 2 + bx + c = (x + p)(x + q), if c is negative, p and q must have opposite signs. (x + p)(x + q), where p and q are positive is x 2 + 5x + 6. It can be factored as (x + 2)(x + 3), where p = 2 and q = 3. Monitoring Progress and Modeling with Mathematics 3. x 2 + 8x + 7
Factors of 7
7. y 2 + 13y − 30
Factors of −30
−1, 1, −2, 2, −3, 3, −5, 30 −30 15 −15 10 −10 6
Sum of factors
29
−29
13
−13
7
−7
1
5, −6 −1
Factors of −42
−1, 1, −2, 2, −3, 3, −6, 42 −42 21 −21 14 −14 7
Sum of factors
41
19
−19
− v − 42 = (v + 6)(v − 7).
11
−11
1
6, −7 −1
1, 7 8
So, x 2 + 8x + 7 = (x + 1)(x + 7). 4. z2 + 10z + 21
Sum of factors
8. v 2 − v − 42
−41
Sum of factors
Factors of 21
So, y 2 + 13y − 30 = (y − 2)(y + 15).
So,
1. The signs tell you that the factors have constant terms
2. Sample answer: A trinomial that can be factored as
6. x 2 + 2x − 15
v2
+ 20
s = 20
Vocabulary and Core Concept Check
Factors of 24
So,
+ 20
0
7.5 Exercises (pp. 389–390)
− 12n + 35 = (n − 5)(n − 7).
5. x 2 − 14x + 24
x2
+ 50
s − 20 =
The diagram shows that the side length is more than 40 meters, so a side length of 20 meters does not make sense in this situation. The side length is 50 meters. So, the area of the square plot of land is 50(50) = 2500 square meters.
4. n2 − 12n + 35
Factors of 35
or
0
s = 50
So, w 2 − 4w + 3 = (w − 1)(w − 3).
n2
− 200
0 = s 2 − 70s + 1000
1, 21
3, 7
22
10
So, z2 + 10z + 21 = (z + 3)(z+ 7). 5. n2 + 9n + 20
Factors of 20 Sum of factors
1, 20
2, 10
4, 5
21
12
9
So, n2 + 9n + 20 = (n + 4)(n + 5). 6. s 2 + 11s + 30
Factors of 30 Sum of factors
1, 30
2, 15
3, 10
5, 6
31
17
13
11
So, s 2 + 11s + 30 = (s + 5)(s + 6).
406
Algebra 1 Worked-Out Solutions
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Chapter 7 7. h 2 + 11h + 18
Factors of 18 Sum of factors
16. z2 + 7z − 18
1, 18
2, 9
3, 6
19
11
18
Factors of −18 Sum of factors
So, h 2 + 11h + 18 = (h + 2)(h + 9). 8. y 2 + 13y + 40
Factors of 40 Sum of factors
1, 40
2, 20
4, 10
5, 8
41
22
14
13
Sum of factors
−1, −4 −2, −2 −5
−1, −22
−2, −11
−23
−13
Sum of factors
−1, −6
−2, −3
5, −8
Sum of factors
−3
39
−39
18
−18
−6
6
3
−7
−5
Factors −1, 1, −2, 2, −3, 3, −4, 4, −6, 6, of −48 48 −48 24 −24 16 −16 12 −12 8 −8 Sum of 47 −47 22 −22 13 −13 8 factors
So, d 2 − 5d + 6 = (d − 2)(d − 3).
Factors of 24 Sum of factors
−1, −24 −2, −12 −3, −8 −25
−14
−4, −6
−11
−10
So, k 2 − 10k + 24 = (k − 4)(k − 6). − 17w + 72 −1, −72
−2, −36
−3, −24
−4, −18
−6, −12
−8, −9
Sum of factors
−73
−38
−27
−22
−18
−17
So, w 2 − 17w + 72 = (w − 8)(w − 9). 14. j 2 − 13j + 42
Sum of factors
2
−2
20. h 2 + 6h − 27
Factors of −27
−1, 27
1, −27
−3, 9
3, −9
Sum of factors
26
−26
6
−6
So, h 2 + 6h − 27 = (h − 3)(h + 9).
Factors of 72
Factors of 42
−8
So, y 2 + 2y − 48 = ( y − 6)( y + 8).
12. k 2 − 10k + 24
21. x 2 − x − 20
Factors of −20
−1, 20
1, −20
−2, 10
2, −10
−4, 5
4, −5
Sum of factors
19
−19
8
−8
1
−1
So, x 2 − x − 20 = (x + 4)(x − 5). −1, −42 −2, −21 −3, −14 −6, −7 −43
−23
−17
So, j 2 − 13j + 42 = ( j − 6)( j − 7). 15.
Factors −1, 1, −2, 2, −4, 4, −5, of −40 40 −40 20 −20 10 −10 8
19. y 2 + 2y − 48
Factors of 6
x2
−1
1
So, s 2 + 3s − 40 = (s − 5)(s + 8).
11. d 2 − 5d + 6
13.
−4
4
18. s 2 + 3s − 40
So, x 2 − 13x + 22 = (x − 2)(x − 11).
w2
−11
11
So, n2 + 4n − 12 = (n − 2)(n + 6).
−4
10. x 2 − 13x + 22
Sum of factors
−3
3
Factors of −12 −1, 12 1, −12 −2, 6 2, −6 −3, 4 3, −4
So, v 2 − 5v + 4 = (v − 1)(v − 4).
Factors of 22
−7
7
17. n2 + 4n − 12
9. v 2 − 5v + 4
Sum of factors
−17
17
So, z2 + 7z − 18 = (z − 2)(z + 9).
So, y 3 + 13y + 40 = (y + 5)(y + 8).
Factors of 4
−1, 18 1, −18 −2, 9 2, −9 −3, 6 3, −6
+ 3x − 4
Factors of −4
−1, 4
1, −4
−2, 2
Sum of factors
3
−3
0
−13
22. m2 − 6m − 7
Factors of −7
−1, 7
1, −7
Sum of factors
6
−6
So, m2 − 6m − 7 = (m + 1)(m − 7).
So, x 2 + 3x − 4 = (x − 1)(x + 4).
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Algebra 1 Worked-Out Solutions
407
Chapter 7 23. −6t − 16 + t 2 = t 2 − 6t − 16
Factors of −16
−1, 16
Sum of factors So,
t2
28. Because c is −60, the factors should have opposite signs.
1, −16
−2, 8
−15
15
2, −8 −6
6
s 2 − 17s − 60
−4, 4
Factors of −60
0
− 6t − 16 = (t + 2)(t − 8). Sum of factors
24. −7y + y 2 − 30 = y 2 − 7y − 30
Factors of −30
−1, 30 1, −30 −3, 10 3, −10 −5, 6 5, −6
Sum of factors
29
−29
−7
7
1
−1
1, −60
−2, 30
2, −30
−3, 20
3, −20
−4, 15
4, −15
−5, 12
5, −12
−6, 10
6, −10
59
−59
28
−28
17
−17
11
−11
7
−7
4
−4
So, s2 − 17s − 60 = (s + 3)(s − 20). 29.
m2 + 3m + 2 = 0 (m + 1)(m + 2) = 0
So, y 2 − 7y − 30 = ( y + 3)( y − 10).
Factors of 2
25. a. x 2 − 8x + 15
Factors of 15
−1, 60
−1, −15
−3, −5
−16
−8
Sum of factors
1, 2
Sum of factors
3
m+1= −1
So, x 2 − 8x + 15 = (x − 3)(x − 5), and a binomial that represents the height of the projection is (x − 5) feet.
−1
0 −2
m = −2
The roots are m = −1 and m = −2.
P = 2ℓ + 2w 30.
= 2(x − 3) + 2(x − 5)
n2 − 9n + 18 = 0
= 2(8 − 3) + 2(8 − 5)
(n − 6)(n − 3) = 0
= 2(5) + 2(3)
Factors of 18
= 10 + 6
−1, −18 −2, −9 −3, −6 −19
Sum of factors
= 16
n−6=
So, the perimeter of the projection is 16 feet.
+6
26. a. x 2 + x − 30
Factors of −30 −1, 30 1, −30 −2, 15 2, −15 −5, 6 5, −6 29
−2
m = −1
b. Let x = 8.
Sum of factors
m+2=
or
0
−29
−13
13
1
−1
So, + x − 30 = (x − 5)(x + 6), and a binomial that represents the width of the land is (x − 5) meters.
−11 n−3=
or
0
−9
+6
+3
n=6
0 +3
n=3
The roots are n = 6 and n = 3. 31. x 2 + 5x − 14 = 0
(x − 2)(x + 7) = 0
x2
Factors of −14
−1, 14
1, −14
−2, 7
2, −7
b. Let x = 20.
Sum of factors
13
−13
5
−5
x 2 + x − 30 = 202 + 20 − 30
x−2=
= 400 + 20 − 30
+2
= 390
0 +2
or
x+7= −7
x=2
0
−7
x = −7
The roots are x = 2 and x = −7.
So, the area of the land is 390 square meters. 27. The sum of the factors 4 and 12 is 16, not 14.
x 2 + 14x + 48 Factors of 48 Sum of factors
1, 48
2, 24
3, 16
4, 12
6, 8
49
26
19
16
14
So, x 2 + 14x + 48 = (x + 6)(x + 8).
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Chapter 7 36.
32. v 2 + 11v − 26 = 0
y 2 − 2y − 8 − 7 = 7 − 7
(v − 2) (v + 13) = 0 Factors of −26
−1, 26
1, −26
−2, 13
2, −13
Sum of factors
25
−25
11
−11
v−2=
v + 13 =
or
0
+2
+2
− 13
v=2
y 2 − 2y − 15 = 0 ( y + 3)( y − 5) = 0 Factors of −15 −1, 15 1, −15 −3, 5 3, −5
0
Sum of factors
− 13
y+3=
v = −13
−3
The roots are v = 2 and v = −13.
(t + 3)(t + 12) = 0 1, 36 2, 18 3, 12 37
t+3=
0
20
15
or t + 12 =
−3
− 12
t = −3
4, 9
6, 6
13
12
m2 − 15m + 10 + 34 = −34 + 34 m2 − 15m + 44 = 0
0
(m − 4)(m − 11) = 0
− 12
m−4= +4
−3, 3, −4, 4, −2, 2, 1, Factors −1, 8 −8 6 24 −24 12 −12 −6 of −24
−3
0
or
−10
10
n−8=
−3
+8
n = −3
−5
5
2
n=8
+ 5 − 8b = 8b − 8b − 10
b 2 − 8b + 5 = −10 b 2 − 8b + 5 + 10 = −10 + 10
+3
0 +3
b=3
−1, −15 −3, −5
or
−16
−8
b−5= +5
0 +5
b=5
The roots are b = 3 and b = 5.
Factors of −50
−1, 50
1, −50
−2, 25
2, −25
−5, 10
5, −10
Sum of factors
49
−49
23
−23
5
−5
a−5=
0 +5
b2
b−3=
(a − 5)(a + 10) = 0
a=5
−2
b 2 + 5 = 8b − 10
38.
Sum of factors
a 2 + 5a − 50 = 0
+5
+ 11
m = 11
Factors of 15
+ 5a − 20 − 30 = 30 − 30
or
0
+ 11
(b − 3)(b − 5) = 0
+8
a 2 + 5a − 20 = 30 a2
m − 11 =
−15
b 2 − 8b + 15 = 0
0
The roots are n = −3 and n = 8. 35.
−24
The roots are m = 4 and m = 11.
(n + 3)(n − 8) = 0
n+3=
or
0 +4
m=4
n2 − 5n − 24 = 0
−23
−45
Sum of factors
n2 − 5n − 24 = 24 − 24
23
−1, −44 −2, −22 −4, −11
Factors of 44
t = −12
n2 − 5n = 24
Sum of factors
y=5
m2 − 15m + 10 = −34
The roots are t = −3 and t = −12. 34.
+5
+ 10 − 15m = 15m − 15m − 34
m2
Sum of factors
0
+5
m2 + 10 = 15m − 34
37.
t 2 + 15t + 36 = 0
−3
−3
−2
The roots are y = −3 and y = 5.
t 2 + 15t + 36 = −36 + 36
Factors of 36
2
y−5=
or
0
−14
14
y = −3
t 2 + 15t = −36
33.
y 2 − 2y − 8 = 7
a + 10 = − 10
0 − 10
a = −10
The roots are a = 5 and a = −10. Copyright © Big Ideas Learning, LLC All rights reserved.
Algebra 1 Worked-Out Solutions
409
Chapter 7 39. The length of the cut picture is (x − 5) inches and the width
is (x − 6) inches.
b.
3 Area of =— 13 browser window
A = ℓw
3 24 = — 13 13 13 — (24) = — 3 3 104 = A
20 = (x − 5)(x − 6) 20 = x(x) + x(−6) + (−5)x + (−5)(−6) 20 =
x2
− 6x − 5x + 30
20 = x 2 − 11x + 30
−1, −10
−2, −5
Sum of factors
−11
−7
x − 10 =
0 or
x−1=
—
+1
x = 10
104 = 13w 104 13w —=— 13 13 8=w So, the screen is 6 + 7 = 13 inches by 8 inches.
0 +1
41. yes; x 2 + bx − 12
x=1
The solutions are x = 10 and x = 1. However, the diagram shows that the side length is more than 6 inches, so a side length of 1 inch does not make sense in this situation. The original picture, therefore, is 10 inches by 10 inches. So, the area of the original square picture is 10(10) = 100 square inches. A = ℓw
40. a.
( 133 ) ⋅ A
104 = (6 + 7)w
0 = (x − 10)(x − 1)
+ 10
⋅A
104 = (x + 7)w
0 = x 2 − 11x + 10
+ 10
Area of screen
Let A = ℓw = (x + 7)w and x = 6.
20 − 20 = x 2 − 11x + 30 − 20
Factors of 10
⋅
24 = x(x − 2) 24 = x(x) + x(−2)
Factors of −12 Sum of factors, b
−1, 12 1, −12 −2, 6 2, −6 −3, 4 3, −4 11
−11
4
−4
1
−1
As shown in the table, there are 6 pairs of integer factors that can be the values of p and q such that (x + p)(x + q) = x 2 + bx − 12. The second row shows the respective corresponding values of b that go with each factor pair. 42. a.
24 = x 2 − 2x 24 − 24 = x 2 − 2x − 24 0 = x 2 − 2x − 24 0 = (x + 4)(x − 6) −3, 3, −4, 4, −2, 2, 1, Factors −1, 6 8 24 −24 12 −12 −6 −8 of −24 Sum of factors
23
x+4= −4
−23 0
−4
x = −4
or
10
−10
x−6= +6
5
−5
2
So, x 2 − x − 6 = x 2 + (2x − 3x) − 6 = (x + 2)(x − 3). b.
−2
0 +6
x=6
So, x2 + 2x − 8 = x2 + (−2x + 4x) − 8 = (x − 2)(x + 4).
The solutions are x = −4 and x = 6, but a length of −4 inches does not make sense in this situation. So, the length of the browser window is 6 inches.
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Chapter 7 A = ℓw
43.
45. To find the equation of the form x 2 + bx + c = 0 that
44 = (x − 5)(x − 12) 44 = x(x) + x(−12) + (−5)(x) + (−5)(−12) 44 = x 2 − 12x − 5x + 60
has solutions x = −4 and x = 6, multiply (x + 4)(x − 6), simplify the polynomial, and substitute the values into the equation. (x + 4)(x − 6) = x(x) + x(−6) + 4(x) + 4(−6)
44 = x 2 − 17x + 60
= x 2 − 6x + 4x − 24
44 − 44 = x 2 − 17x + 60 − 44
= x 2 − 2x − 24
0 = x 2 − 17x + 16
So, the equation x 2 − 2x − 24 = 0 has solutions x = −4 and x = 6.
0 = (x − 1)(x − 16) Factors of 16
−1, −16
−2, −8
−4, 4
−17
−10
0
Sum of factors x−1= +1
0 or +1
x − 16 = + 16
x=1
0
46. a. Look for the x-intercepts, and substitute them for p and q
in (x − p)(x − q). b. The graph crosses the x-axis at x = −3 and x = 2.
So, x 2 + x − 6 = (x + 3)(x − 2).
+ 16
x = 16
The solutions are x = 1 and x = 16. However, a side length of x − 5 = 1 − 5 = −4 feet does not make sense in this situation. So, use the solution x = 16. Therefore, the rectangle is x − 5 = 16 − 5 = 11 feet by x − 12 = 16 − 12 = 4 feet.
47. a. Area being
=
paved
A = ℓw + ℓw = 20(x) + x(18 − x) = 20x + x(18) + x(−x)
A = —12 bh
44.
= 20x + 18x − x 2 = −x 2 + 38x
35 = —12 (g − 8)(g − 11) 35 = —12 [ g(g) + g(−11) + (−8)(g) + (−8)(−11) ] 35 = —12 ( g2 − 11g − 8g + 88 )
⋅
2 35 = 2
⋅
(
g2
An expression that represents the area being paved is ( −x 2 + 38x ) square meters. b. A = −x 2 + 38x
35 = —12 ( g2 − 19g + 88 ) 1 —2
An equation that can be used to find the width of the road is 280 = −x 2 + 38x.
− 19g + 88 )
280 = −x 2 + 38x
70 = g2 − 19g + 88
280 + x 2 = −x 2 + x 2 + 38x
70 − 70 = g2 − 19g + 88 − 70
x 2 + 280 = 38x
0 = g2 − 19g + 18
x 2 + 280 − 38x = 38x − 38x
0 = (g − 1)(g − 18) Factors of 18
−1, −18
−2, −9
−3, −6
−19
−11
−9
Sum of factors g−1= +1
0 +1
g=1
Area of bottom Area of top + rectangle rectangle
or
g − 18 = + 18
0 + 18
g = 18
The solutions are g = 1 and g = 18. However, a side length of g − 8 = 1 − 8 = −7 meters does not make sense in this situation. So, use the solution g = 18. Therefore, the triangle has a base that is g − 8 = 18 − 8 = 10 meters and a height that is g − 11 = 18 − 11 = 7 meters.
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x 2 − 38x + 280 = 0 (x − 10)(x − 28) = 0 x − 10 = + 10
0 + 10
x = 10
or
x − 28 = + 28
0 + 28
x = 28
Factors −1, −2, −4, −5, −7, −8, −10, −14, of 280 −280 −140 −70 −56 −40 −35 −28 −20 Sum of factors −281 −142 −74 −61 −47 −43 −38 −34 The solutions are x = 10 and x = 28. However, the diagram shows that the width of the road is less than 18 meters, so a width of 28 meters does not make sense in this situation. Therefore, the width of the road is 10 meters.
Algebra 1 Worked-Out Solutions
411
Chapter 7 7.6 Explorations (p. 391)
48. x 2 + 6xy + 8y 2
Factors of 8
1. a.
1, 8 2, 4
Sum of factors
9
6
So, x 2 + 6xy + 8y 2 = (x + 2y)(x + 4y). 49. r 2 + 7rs + 12s 2
Factors of 12
So, 3x 2 + 5x + 2 = (x + 1)(3x + 2). 1, 12 2, 6 3, 4
Sum of factors
13
8
b.
7
So, r 2 + 7rs + 12s 2 = (r + 3s)(r + 4s). 50. a 2 + 11ab − 26b 2
Factors of −26 −1, 26 1, −26 −2, 13 2, −13 Sum of factors
−25
25
11
−11
So, a 2 + 11ab − 26b 2 = (a − 2b)(a + 13b).
So, 4x 2 + 4x − 3 = (2x + 3)(2x − 1). c.
51. x 2 − 2xy − 35y 2
Factors of −35 −1, 35 1, −35 Sum of factors
−5, 7
5, −7
2
−2
−34
34
So, x 2 − 2xy − 35y 2 = (x + 5y)(x − 7y). Maintaining Mathematical Proficiency 52. p − 9 = 0
Check p − 9 = 0 ? 9−9=0
+9 +9 p=9
0=0✓
The solution is p = 9. 53. z + 12 = −5
− 12
Check
− 12
z + 12 = −5 ? −17 + 12 = −5
z = −17
−5 = −5 ✓
So, 2x 2 − 11x + 5 = (x − 5)(2x − 1). 2. Arrange algebra tiles that model the trinomial into a
rectangular array, use additional algebra tiles to model the dimensions of the rectangle, then write the polynomial in factored form using the dimensions of the rectangle. 3. The trinomial 2x 2 + 2x + 1 cannot be factored because the
expression cannot be modeled as a rectangular array.
The solution is z = −17. 7.6 Monitoring Progress (pp. 393–394) 54.
c 6=— −7 c −7(6) = −7 — −7 −42 = c
( )
c Check 6 = — −7 ? −42 6=— −7 6=6✓
The solution is c = −42. 55. 4k = 0
4k 0 —=— 4 4 k=0
1. 8x 2 − 56x + 48 = 8( x 2 − 7x + 6 )
= 8(x − 6)(x − 1) Factors of 6 Sum of factors
−1, −6 −2, −3 −7
−5
So, 8x 2 − 56x + 48 = 8(x − 6)(x − 1).
Check 4k = 0 ? 4(0) = 0 0=0✓
The solution is k = 0.
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Chapter 7 2. 14x 2 + 31x + 15
4. 3x 2 − 14x + 8
Factors Factors Possible of 14 of 15 factorization 1, 14
1, 15
(x + 1) (14x + 15)
1, 14
15, 1
(x + 15) (14x + 1)
x + 210x = 211x
1, 14
3, 5
(x + 3) (14x + 5)
5x + 42x = 47x
1, 14
5, 3
(x + 5) (14x + 3)
3x + 70x = 73x
2, 7
1, 15
(2x + 1) (7x + 15)
30x + 7x = 37x
2, 7
15, 1
(2x + 15) (7x + 1)
2x + 105x = 107x
2, 7
3, 5
(2x + 3) (7x + 5)
10x + 21x = 31x
(2x + 5) (7x + 3)
6x + 35x = 41x
2, 7 So,
14x 2
5, 3
Factors Factors Possible of 3 of 8 factorization
Middle term 15x + 14x = 29x
✗
1, 3
−1, −8
(x − 1) (3x − 8)
−8x − 3x = −11x
✗
1, 3
−8, −1
(x − 8) (3x − 1)
−x − 24x = −25x
✗
1, 3
−2, −4
(x − 2) (3x − 4)
−4x − 6x = −10x
✗
1, 3
−4, −2
(x − 4) (3x − 2)
−2x − 12x = −14x
✗ ✗ ✓
1, 2 1, 2
−1, −5 −5, −1
Middle term
(x − 1) (2x − 5)
−5x − 2x = −7x
(x − 5) (2x − 1)
−x − 10x = −11x
So, 2x 2 − 7x + 5 = (x − 1)(2x − 5).
✓ ✗
✗ ✗ ✓
5. 4x 2 − 19x − 5
Factors Factors Possible of 4 of −5 factorization
Middle term
1, 4
1, −5
(x + 1) (4x − 5)
−5x + 4x = −x
✗
1, 4
5, −1
(x + 5) (4x − 1)
−x + 20x = 19x
✗
1, 4
−1, 5
(x − 1) (4x + 5)
5x − 4x = x
✗
1, 4
−5, 1
(x − 5) (4x + 1)
x − 20x = −19x ✓
2, 2
1, −5
(2x + 1) (2x − 5)
−10x + 2x = −8x
✗
2, 2
−1, 5
(2x − 1) (2x + 5)
10x − 2x = 8x
✗
✗
3. 2x 2 − 7x + 5
✗
So, 3x 2 − 14x + 8 = (x − 4)(3x − 2).
+ 31x + 15 = (2x + 3)(7x + 5).
Factors Factors Possible of 2 of 5 factorization
Middle term
So, 4x 2 − 19x − 5 = (x − 5)(4x + 1).
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Algebra 1 Worked-Out Solutions
413
Chapter 7 6. 6x 2 + x − 12
Factors Factors Possible of 6 of −12 factorization Middle term 1, −12 (x + 1) −12x + 6x = −6x 1, 6 ✗ (6x − 12) −x + 72x = 71x 12, −1 1, 6 (x + 12) ✗ (6x − 1) 12x − 6x = 6x −1, 12 1, 6 (x − 1) ✗ (6x + 12) x − 72x = −71x −12, 1 1, 6 (x − 12) ✗ (6x + 1) −6x + 12x = 6x 2, −6 1, 6 (x + 2) ✗ (6x − 6) −2x + 36x = 34x 6, −2 1, 6 (x + 6) ✗ (6x − 2) 6x − 12x = −6x −2, 6 1, 6 (x − 2) ✗ (6x + 6) 2x − 36x = −34x −6, 2 1, 6 (x − 6) ✗ (6x + 2) −4x + 18x = 14x 3, −4 1, 6 (x + 3) ✗ (6x − 4) 4, −3 −3x + 24x = 21x 1, 6 (x + 4) ✗ (6x − 3) 4x − 18x = −14x −3, 4 1, 6 (x − 3) ✗ (6x + 4) −4, 3 3x − 24x = −21x 1, 6 (x − 4) ✗ (6x + 3) −24x + 3x = −21x 1, −12 2, 3 (2x + 1) ✗ (3x − 12) 12, −1 2, 3 (2x + 12) −2x + 36x = 34x ✗ (3x − 1) 24x − 3x = 21x −1, 12 2, 3 (2x − 1) ✗ (3x + 12) −12, 1 2x − 36x = −34x 2, 3 (2x − 12) ✗ (3x + 1) −12x + 6x = −6x 2, −6 2, 3 (2x + 2) ✗ (3x − 6) −4x + 18x = 14x 6, −2 2, 3 (2x + 6) ✗ (3x − 2) 12x − 6x = 6x −2, 6 2, 3 (2x − 2) ✗ (3x + 6) −6, 2 4x − 18x = −14x 2, 3 (2x − 6) ✗ (3x + 2) −8x + 9x = x 3, −4 2, 3 (2x + 3) ✓ (3x − 4) −6x + 12x = 6x 4, −3 2, 3 (2x + 4) ✗ (3x − 3) 8x − 9x = −x −3, 4 2, 3 (2x − 3) ✗ (3x + 4) −4, 3 6x − 12x = −6x 2, 3 (2x − 4) ✗ (3x + 3)
7. −2y 2 − 5y − 3 = −( 2y 2 + 5y + 3 )
Factors Factors Possible of 2 of 3 factorization
Middle term
1, 2
1, 3
( y + 1) (2y + 3)
3y + 2y = 5y
1, 2
3, 1
( y + 3) (2y + 1)
y + 6y = 7y
✓ ✗
So, −2y 2 − 5y − 3 = −( y + 1)(2y + 3). 8. −5m2 + 6m − 1 = −( 5m2 − 6m + 1 )
Factors Factors Possible of 5 of 1 factorization 1, 5
−1, −1
(m − 1) (5m − 1)
Middle term −m − 5m = −6m
✓
So, −5m2 + 6m − 1 = −(m − 1)(5m − 1). 9. −3x 2 − x + 2 = −( 3x 2 + x − 2 )
Factors Factors Possible of 3 of −2 factorization 1, 3
1, −2
(x + 1) (3x − 2)
1, 3
2, −1
(x + 2) (3x − 1)
1, 3
−1, 2
(x − 1) (3x + 2)
1, 3
−2, 1
(x − 2) (3x + 1)
Middle term −2x + 3x = x −x + 6x = 5x 2x − 3x = −x x − 6x = −5x
✓ ✗ ✗ ✗
So, −3x 2 − x + 2 = −(x + 1)(3x − 2).
So, 6x 2 + x − 12 = (2x + 3)(3x − 4).
414
Algebra 1 Worked-Out Solutions
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Chapter 7 10.
7.6 Exercises (pp. 395–396)
w(2w + 1) = 136 w(2w) + w(1) = 136
Vocabulary and Core Concept Check
2w2 + w = 136
1. The greatest common factor of the terms of
3y 2 − 21y + 36 is 3.
2w2 + w − 136 = 136 − 136 2w2 + w − 136 = 0
2. Factoring 6x 2 − x − 2 requires considering factors of 6 and
Factors Factors Possible of 2 of −136 factorization
Middle term
1, 2
1, −136
(w + 1) −136w + 2w = −134w ✗ (2w − 136)
1, 2
136, −1
(w + 136) (2w − 1)
−w + 272w = 271w
1, 2
−1, 136
(w − 1) (2w + 136)
136w − 2w = 134w
1, 2
−136, 1
(w − 136) (2w + 1)
1, 2
2, −68
(w + 2) (2w − 68)
w − 272w = −271w −68w + 4w = −64w
1, 2
68, −2
(w + 68) (2w − 2)
−2w + 136w = 134w
1, 2
−2, 68
(w − 2) (2w + 68)
1, 2
−68, 2
(w − 68) (2w + 2)
1, 2
4, −34
(w + 4) (2w − 34)
−34w + 8w = −26w
1, 2
34, −4
(w + 34) (2w − 4)
−4w + 68w = 64w
68w − 4w = 64w 2w − 136w = −134w
1, 2
−4, 34
(w − 4) (2w + 34)
34w − 8w = 26w
1, 2
−34, 4
(w − 34) (2w + 4)
4w − 68w = −64w
1, 2
8, −17
(w + 8) (2w − 17)
−17w + 16w = −w
1, 2
17, −8
(w + 17) (2w − 8)
−8w + 34w = 26w
1, 2
−8, 17
(w − 8) (2w + 17)
−17, 8
(w − 17) (2w + 8)
1, 2
17w − 16w = w 8w − 34w = −26w
✗ ✗ ✗ ✗
or
Monitoring Progress and Modeling with Mathematics 3. 3x 2 + 3x − 6 = 3( x 2 + x − 2 )
= 3(x − 1)(x + 2) Factors of −2
−1, 2
1, −2
Sum of factors
1
−1
So, 3x 2 + 3x − 3 = 3(x − 1)(x + 2). 4. 8v 2 + 8v − 48 = 8( v 2 + v − 6 )
= 8(v − 2)(v + 3)
✗ ✗ ✗ ✗ ✗
Factors of −6
−1, 6
1, −6
−2, 3
2, −3
Sum of factors
5
−5
1
−1
So, 8v 2 + 8v − 48 = 8(v − 2)(v + 3). 5. 4k 2 + 28k + 48 = 4( k 2 + 7k + 12 )
= 4(k + 3)(k + 4) Factors of 12 Sum of factors
✗ ✗
✗ ✓ ✗
2w + 17 = 0 −17 −17
2w = −17 2w −17 —=— 2 2 17 w = −— 2 A negative width does not make sense, so you should use the positive solution. So, the width of the reserve is 8 miles.
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2, 6
3, 4
13
8
7
So, 4k 2 + 28k + 48 = 4(k + 3)(k + 4). 6. 6y 2 − 24y + 18 = 6( y 2 − 4y + 3 )
−1, −3
Factors of 3 Sum of factors
−4
So, 6y 2 − 24y + 18 = 6(y − 1)(y − 3). 7. 7b 2 − 63b + 140 = 7( b 2 − 9b + 20 )
= 7(b − 4)(b − 5) Factors of 20 Sum of factors
w=8
1, 12
= 6( y − 1)(y − 3) ✗
(w − 8)(2w + 17) = 0 w−8= 0 +8 +8
−2 in different combinations until the combination is found that produces the correct middle term. Factoring x 2 − x − 2 only requires finding the factors of −2 that add up to −1.
−1, −20
−2, −10
−4, −5
−22
−12
−9
So, 7b 2 − 63b + 140 = 7(b − 4)(b − 5). 8. 9r 2 − 36r − 45 = 9( r 2 − 4r − 5 )
= 9(r + 1)(r − 5) Factors of −5
1, −5
−1, 5
Sum of factors
−4
4
So, 9r 2 − 36r − 45 = 9(r + 1)(r − 5).
Algebra 1 Worked-Out Solutions
415
Chapter 7 9. 3h2 + 11h + 6
12. 10w2 − 31w + 15
Factors Factors Possible of 3 of 6 factorization
Factors Factors of 10 of 15
Middle term
1, 3
1, 6
(h + 1) (3h + 6)
6h + 3h = 9h
1, 3
6, 1
(h + 6) (3h + 1)
h + 18h = 19h
1, 3
2, 3
(h + 2) (3h + 3)
3h + 6h = 9h
1, 3
3, 2
(h + 3) (3h + 2)
2h + 9h = 11h
1, 10
−1, −15
(w − 1) (10w − 15)
✗
1, 10
−15, −1
(w − 15) (10w − 1)
−w − 150w = −151w ✗
✗
1, 10
−3, −5
(w − 3) (10w − 5)
−5w − 30w = −35w
✗
✓
1, 10
−5, −3
(w − 5) (10w − 3)
−3w − 50w = −53w
✗
2, 5
−1, −15
(2w − 1) (5w − 15)
−30w − 5w = −35w
✗
2, 5
−15, −1
(2w − 15) (5w − 1)
−2w − 75w = −77w
✗
2, 5
−3, −5
(2w − 3) (5w − 5)
−10w − 15w = −25w
✗
2, 5
−5, −3
(2w − 5) (5w − 3)
−6w − 25w = −31w
✓
10. 8m2 + 30m + 7
1, 8
1, 7
Middle term
(m + 1) (8m + 7)
7m + 8m = 15m
1, 8
7, 1
(m + 7) (8m + 1)
m + 56m = 57m
2, 4
1, 7
(2m + 1) (4m + 7)
14m + 4m = 18m
2, 4
7, 1
(2m + 7) (4m + 1)
2m + 28m = 30m
✗ ✗ ✗ ✓
So, 8m2 + 30m + 7 = (2m + 7)(4m + 1). 11. 6x 2 − 5x + 1
Factors Factors Possible of 6 of 1 factorization
Middle term
✗
So, 3h2 + 11h + 6 = (h + 3)(3h + 2).
Factors Factors Possible of 8 of 7 factorization
Possible factorization
Middle term
1, 6
−1, −1
(x − 1) (6x − 1)
−x − 6x = −7x
✗
2, 3
−1, −1
(2x − 1) (3x − 1)
−2x − 3x = −5x
✓
−15w − 10w = −25w
✗
So, 10w2 − 31w + 15 = (2w − 5)(5w − 3) 13. 3n2 + 5n − 2
Factors Factors Possible of 3 of −2 factorization 1, 3
1, −2
(n + 1) (3n − 2)
1, 3
2, −1
(n + 2) (3n − 1)
1, 3
−1, 2
(n − 1) (3n + 2)
1, 3
−2, 1
(n − 2) (3n + 1)
Middle term −2n + 3n = n −n + 6n = 5n 2n − 3n = −n n − 6n = −5n
✗ ✓ ✗ ✗
So, 3n2 + 5n − 2 = (n + 2)(3n − 1).
So, 6x 2 − 5x + 1 = (2x − 1)(3x − 1).
416
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Chapter 7 14. 4z2 + 4z − 3
16. 18v2 − 15v − 18 = 3(6v2 − 5v − 6)
Factors Factors Possible of 4 of −3 factorization
Factors Factors Possible of 6 of −6 factorization
Middle term
Middle term
1, 4
1, −3
(z + 1) (4z − 3)
−3z + 4z = z
✗
1, 6
1, −6
(v + 1) (6v − 6)
−6v + 6v = 0
✗
1, 4
3, −1
(z + 3) (4z − 1)
−z + 12z = 11z
✗
1, 6
6, −1
(v + 6) (6v − 1)
−v + 36v = 35v
✗
1, 4
−1, 3
(z − 1) (4z + 3)
3z − 4z = −z
✗
1, 6
−1, 6
(v − 1) (6v + 6)
6v − 6v = 0
✗
1, 4
−3, 1
(z − 3) (4z + 1)
z − 12z = −11z
✗
1, 6
−6, 1
(v − 6) (6v + 1)
v − 36v = −35v ✗
2, 2
1, −3
(2z + 1) (2z − 3)
−6z + 2z = −4z
✗
1, 6
2, −3
(v + 2) (6v − 3)
−3v + 12v = 9v
✗
2, 2
−1, 3
(2z − 1) (2z + 3)
6z − 2z = 4z
✓
1, 6
3, −2
(v + 3) (6v − 2)
−2v + 18v = 16v
✗
1, 6
−2, 3
(v − 2) (6v + 3)
3v − 12v = −9v
1, 6
−3, 2
(v − 3) (6v + 2)
2v − 18v = −16v ✗
2, 3
1, −6
(2v + 1) (3v − 6)
−12v + 3v = −9v
✗
2, 3
6, −1
(2v + 6) (3v − 1)
−2v + 18v = 16v
✗
2, 3
−1, 6
(2v − 1) (3v + 6)
12v − 3v = 9v
✗
2, 3
−6, 1
(2v − 6) (3v + 1)
2v − 18v = −16v ✗
2, 3
2, −3
(2v + 2) (3v − 3)
−6v + 6v = 0
✗
2, 3
3, −2
(2v + 3) (3v − 2)
−4v + 9v = 5v
✗
2, 3
−2, 3
(2v − 2) (3v + 3)
6v − 6v = 0
✗
2, 3
−3, 2
(2v − 3) (3v + 2)
4v − 9v = −5v
✓
So, 4z2 + 4z − 3 = (2z − 1)(2z + 3). 15. 8g2 − 10g − 12 = 2(4g2 − 5g − 6)
Factors Factors Possible of 4 of −6 factorization
Middle term
1, 4
1, −6
(g + 1) (4g − 6)
−6g + 4g = −2g
✗
1, 4
6, −1
(g + 6) (4g − 1)
−g + 24g = 23g
✗
1, 4
−1, 6
(g − 1) (4g + 6)
6g − 4g = 2g
1, 4
−6, 1
(g − 6) (4g + 1)
g − 24g = −23g ✗
1, 4
2, −3
(g + 2) (4g − 3)
−3g + 8g = 5g −2g + 12g = 10g
✗
✗
1, 4
3, −2
(g + 3) (4g − 2)
1, 4
−2, 3
(g − 2) (4g + 3)
1, 4
−3, 2
(g − 3) (4g + 2)
2g − 12g = −10g ✗
2, 2
1, −6
(2g + 1) (2g − 6)
−12g + 2g = −10g ✗
2, 2
−1, 6
(2g − 1) (2g + 6)
12g − 2g = 10g
✗
2, 2
2, −3
(2g + 2) (2g − 3)
−6g + 4g = −2g
✗
2, 2
−2, 3
(2g − 2) (2g + 3)
6g − 4g = 2g
✗
3g − 8g = −5g
✗ ✓
✗
So, 18v2 − 15v − 18 = 3(2v − 3)(3v + 2).
So, 8g2 − 10g − 12 = 2(g − 2)(4g + 3).
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Algebra 1 Worked-Out Solutions
417
Chapter 7 17. −3t2 + 11t − 6 = −(3t2 − 11t + 6)
Factors Factors Possible of 3 of −11 factorization 1, 3 −1, −6 (t − 1) (3t − 6) 1, 3 −6, −1 (t − 6) (3t − 1)
20. −8h2 − 13h + 6 = −( 8h2 + 13h − 6 )
Factors Factors Possible of 8 of −6 factorization
Middle term
Middle term
−6t − 3t = −9t
✗
1, 8
1, −6
(h + 1) (8h − 6)
−6h + 8h = 2h
✗
−t − 18t = −19t
✗
1, 8
6, −1
(h + 6) (8h − 1)
−h + 48h = 47h
✗
1, 3
−2, −3
(t − 2) (3t − 3)
−3t − 6t = −9t
✗
1, 8
−1, 6
(h − 1) (8h + 6)
6h − 8h = −2h
1, 3
−3, −2
(t − 3) (3t − 2)
−2t − 9t = −11t
✓
1, 8
−6, 1
(h − 6) (8h + 1)
h − 48h = −47h ✗
1, 8
2, −3
(h + 2) (8h − 3)
−3h + 16h = 13h
✓
1, 8
3, −2
(h + 3) (8h − 2)
−2h + 24h = 22h
✗
1, 8
−2, 3
(h − 2) (8h + 3)
3h − 16h = −13h ✗
1, 8
−3, 2
(h − 3) (8h + 2)
2h − 24h = −22h ✗
2, 4
1, −6
(2h + 1) (4h − 6)
−12h + 4h = −8h
✗
2, 4
6, −1
(2h + 6) (4h − 1)
−2h + 24h = 22h
✗
2, 4
−1, 6
(2h − 1) (4h + 6)
12h − 4h = 8h
✗
2, 4
−6, 1
(2h − 6) (4h + 1)
2h − 24h = −22h ✗
2, 4
2, −3
(2h + 2) (4h − 3)
−6h + 8h = 2h
✗
2, 4
3, −2
(2h + 3) (4h − 2)
−4h + 12h = 8h
✗
2, 4
−2, 3
(2h − 2) (4h + 3)
6h − 8h = −2h
✗
2, 4
−3, 2
(2h − 3) (4h + 2)
4h − 12h = −8h
✗
So, −3t2 + 11t − 6 = −(t − 3)(3t − 2). 18. −7v2 − 25v − 12 = −(7v2 + 25v + 12)
Factors Factors Possible of 7 of 12 factorization 1, 7 1, 12 (v + 1) (7v + 12) 1, 7 1, 7 1, 7 1, 7 1, 7
12, 1 2, 6 6, 2 3, 4 4, 3
Middle term 12v + 7v = 19v
✗
(v + 12) (7v + 1)
v + 84v = 85v
✗
(v + 2) (7v + 6)
6v + 14v = 20v
✗
(v + 6) (7v + 2)
2v + 42v = 44v
✗
(v + 3) (7v + 4)
4v + 21v = 25v
✓
(v + 4) (7v + 3)
3v + 28v = 31v
✗
So, −7v2 − 25v − 12 = −(v + 3)(7v + 4). 19. −4c2 + 19c + 5 = −(4c2 − 19c − 5)
Factors Factors Possible of 4 of −5 factorization Middle term 1, 4 ✗ 1, −5 (c + 1) −5c + 4c = −c (4c − 5) 1, 4 ✗ 5, −1 (c + 5) −c + 20c = 19c (4c − 1) 1, 4 ✗ −1, 5 (c − 1) 5c − 4c = c (4c + 5) 1, 4 −5, 1 (c − 5) c − 20c = −19c ✓ (4c + 1) 2, 2 ✗ 1, −5 (2c + 1) −10c + 2c = −8c (2c − 5) 2, 2 ✗ −1, 5 (2c − 1) 10c − 2c = 8c (2c + 5)
✗
So, −8h2 − 13h + 6 = −(h + 2)(8h − 3).
So, −4c2 + 19c + 5 = −(c − 5)(4c + 1).
418
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Chapter 7 21. −15w2 − w + 28 = −(15w2 + w − 28)
Factors Factors Possible of 15 of −28 factorization Middle term 1, 15 1, −28 (w + 1) −28w + 15w = −13w (15w − 28) 1, 15 28, −1 (w + 28) −w + 420w = 419w (15w − 1) 1, 15 −1, 28 (w − 1) 28w − 15w = 13w (15w + 28) 1, 15 −28, 1 (w − 28) w − 420w = −419w (15w + 1) 1, 15 2, −14 (w + 2) −14w + 30w = 16w (15w − 14) 1, 15 14, −2 (w + 14) −2w + 210w = 208w (15w − 2) 1, 15 −2, 14 (w − 2) 14w − 30w = −16w (15w + 14) 1, 15 −14, 2 (w − 14) 2w − 210w = −208w (15w + 2) 1, 15 4, −7 (w + 4) −7w + 60w = 53w (15w − 7) 1, 15 7, −4 (w + 7) −4w + 105w = 101w (15w − 4) 1, 15 −4, 7 (w − 4) 7w − 60w = −53w (15w + 7) 1, 15 −7, 4 (w − 7) 4w − 105w = −101w (15w + 4) 3, 5 1, −28 (3w + 1) −84w + 5w = −79w (5w − 28) 3, 5 28, −1 (3w + 28) −3w + 140w = 137w (5w − 1) 3, 5 −1, 28 (3w − 1) 84w − 5w = 79w (5w + 28) 3, 5 −28, 1 (3w − 28) 3w − 140w = −137w (5w + 1) 3, 5 2, −14 (3w + 2) −42w + 10w = −32w (5w − 14) 3, 5 14, −2 (3w + 14) −6w + 70w = 64w (5w − 2) 3, 5 −2, 14 (3w − 2) 42w − 10w = 32w (5w + 14) 3, 5 −14, 2 (3w − 14) 6w − 70w = −64w (5w + 2) 3, 5 4, −7 (3w + 4) −21w + 20w = −w (5w − 7) 3, 5 7, −4 (3w + 7) −12w + 35w = 23w (5w − 4) 3, 5 −4, 7 (3w − 4) 21w − 20w = w (5w + 7) 3, 5 −7, 4 (3w − 7) 12w − 35w = −23w (5w + 4)
22. −22d 2 + 29d − 9 = −(22d 2 − 29d + 9)
Factors Factors Possible of 22 of 9 factorization
Middle term
✗
1, 22
−1, −9
(d − 1) (22d − 9)
−9d − 22d = −31d
✗
1, 22
−9, −1
(d − 9) (22d − 1)
−d − 198d = −199d ✗
✗
1, 22
−3, −3
(d − 3) (22d − 3)
−3d − 66d = −69d
✗
✗
2, 11
−1, −9
−18d − 11d = −29d
✓
✗
(2d − 1) (11d − 9)
2, 11
−9, −1
−2d − 99d = −101d ✗
✗
(2d − 9) (11d − 1)
2, 11
−3, −3
(2d − 3) (11d − 3)
−6d − 33d = −39d
✗ ✗ ✗ ✗ ✗ ✗
✗
✗
So, −22d 2 + 29d − 9 = −(2d − 1)(11d − 9). 23. The greatest common factor 2 should be factored out of
every term, but it was only factored out of the first term. 2x 2 − 2x − 24 = 2(x 2 − x − 12) = 2(x + 3)(x − 4) Factors −1, 12 1, −12 of −12 Sum of −11 11 factors
−2, 6
2, −6
−3, 4
3, −4
4
−4
1
−1
So, 2x 2 − 2x − 24 = 2(x + 3)(x − 4). ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✓ ✗
So, −15w2 − w + 28 = −(3w − 4)(5w + 7). Copyright © Big Ideas Learning, LLC All rights reserved.
Algebra 1 Worked-Out Solutions
419
Chapter 7 24. These factors do not produce the correct middle term when
26. 2k2 − 5k − 18 = 0
they are multiplied.
Factors Factors Possible of 2 of −18 factorization
6x2 − 7x − 3 Factors Factors Possible of 6 of −3 factorization
(k + 1) (2k − 18)
−18k + 2k = −16k ✗
1, −3
(x + 1) (6x − 3)
−3x + 6x = 3x
✗
1, 2
18, −1
(k + 18) (2k − 1)
−k + 36k = 35k
✗
1, 6
3, −1
(x + 3) (6x − 1)
−x + 18x = 17x
✗
1, 2
−1, 18
(k − 1) (2k + 18)
18k − 2k = 16k
✗
1, 6
−1, 3
(x − 1) (6x + 3)
3x − 6x = −3x
✗
1, 2
−18, 1
(k − 18) (2k + 1)
1, 6
−3, 1
(x − 3) (6x + 1)
x − 18x = −17x
✗
1, 2
2, −9
(k + 2) (2k − 9)
2, 3
1, −3
(2x + 1) (3x − 3)
−6x + 3x = −3x
✗
1, 2
9, −2
(k + 9) (2k − 2)
2, 3
3, −1
(2x + 3) (3x − 1)
−2x + 9x = 7x
✗
1, 2
−2, 9
(k − 2) (2k + 9)
2, 3
−1, 3
(2x − 1) (3x + 3)
6x − 3x = 3x
✗
1, 2
−9, 2
(k − 9) (2k + 2)
2, 3
−3, 1
(2x − 3) (3x + 1)
2x − 9x = −7x
✓
1, 2
3, −6
(k + 3) (2k − 6)
−6k + 6k = 0
✗
1, 2
6, −3
(k + 6) (2k − 3)
−3k + 12k = 9k
✗
1, 2
−3, 6
(k − 3) (2k + 6)
1, 2
−6, 3
(k − 6) (2k + 3)
5x 2 − 5x − 30 = 0
25.
5(x 2
⋅
− x − 6) = 0
5(x + 2)(x − 3) = 0 1 5(x + 2)(x − 3) = — 5 (x + 2)(x − 3) = 0
x+2= −2
0
or
⋅0
−2
k − 36k = −35k ✗ −9k + 4k = −5k −2k + 18k = 16k 9k − 4k = 5k
✓ ✗ ✗
2k − 18k = −16k ✗
6k − 6k = 0 3k − 12k = −9k
✗ ✗
(k + 2)(2k − 9) = 0
x−3= +3
x = −2
k+2=
0
−2
+3
Factors of −6
−1, 6
1, −6
−2, 3
2, −3
Sum of factors
5
−5
1
−1
Algebra 1 Worked-Out Solutions
0 −2
or
2k − 9 = +9
0 +9
k = −2
x=3
The roots are x = −2 and x = 3.
420
1, −18
1, 6
So, 6x 2 − 7x − 3 = (2x − 3)(3x + 1).
1 — 5
1, 2 Middle term
Middle term
2k = 9 2k 9 —=— 2 2 9 k=— 2 9 The roots are k = −2 and k = —. 2
Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 27.
−12n2 − 11n = −15
−(3n + 5)(4n − 3) = 0
−12n2 − 11n + 15 = −15 + 15
−1[ −(3n + 5)(4n − 3) ] = −1(0)
−12n2 − 11n + 15 = 0
(3n + 5)(4n − 3) = 0
−(12n2 + 11n − 15) = 0 Factors Factors Possible of 12 of −15 factorization Middle term 1, −15 (n + 1) −15n + 12n = −3n 1, 12 (12n − 15) 15, −1 (n + 15) −n + 180n = 179n 1, 12 (12n − 1) −1, 15 (n − 1) 15n − 12n = 3n 1, 12 (12n + 15) −15, 1 (n − 15) n − 180n = −179n 1, 12 (12n + 1) 3, −5 (n + 3) −5n + 36n = 31n 1, 12 (12n − 5) 5, −3 (n + 5) −3n + 60n = 57n 1, 12 (12n − 3) −3, 5 (n − 3) 5n − 36n = −31n 1, 12 (12n + 5) −5, 3 (n − 5) 3n − 60n = −57n 1, 12 (12n + 3) 1, −15 (2n + 1) −30n + 6n = −24n 2, 6 (6n − 15) 15, −1 (2n + 15) −2n + 90n = 88n 2, 6 (6n − 1) −1, 15 (2n − 1) 30n − 6n = 24n 2, 6 (6n + 15) −15, 1 (2n − 15) 2n − 90n = −88n 2, 6 (6n + 1) 3, −5 (2n + 3) −10n + 18n = 3n 2, 6 (6n − 5) 5, −3 (2n + 5) −6n + 30n = 24n 2, 6 (6n − 3) −3, 5 (2n − 3) 10n − 18n = −8n 2, 6 (6n + 5) −5, 3 (2n − 5) 6n − 30n = −24n 2, 6 (6n + 3) 1, −15 (3n + 1) −45n + 4n = −41n 3, 4 (4n − 15) 15, −1 (3n + 15) −3n + 60n = 57n 3, 4 (4n − 1) −1, 15 (3n − 1) 45n − 4n = 41n 3, 4 (4n + 15) −15, 1 (3n − 15) 3n − 60n = −57n 3, 4 (4n + 1) 3, −5 (3n + 3) −15n + 12n = −3n 3, 4 (4n − 5) 5, −3 (3n + 5) −9n + 20n = 11n 3, 4 (4n − 3) 15n − 12n = 3n −3, 5 (3n − 3) 3, 4 (4n + 5) −5, 3 (3n − 5) 9n − 20n = −11n 3, 4 (4n + 3) Copyright © Big Ideas Learning, LLC All rights reserved.
3n + 5 = −5
✗
✗
−5
+3
0 +3
4n = 3 4n 3 —=— 4 4 3 n=— 4 5 3 The roots are n = −— and n = —. 3 4
✗
✗
4n − 3 =
or
3n = −5 3n −5 —=— 3 3 5 n = −— 3
✗
✗
0
14b 2 − 2 = −3b
28.
14b 2
− 2 + 3b = −3b + 3b
14b 2
+ 3b − 2 = 0
Factors Factors Possible of 14 of −2 factorization
Middle term
✗
1, 14
1, −2
(b + 1) (14b − 2)
−2b + 14b = 12b
✗
✗
1, 14
2, −1
(b + 2) (14b − 1)
−b + 28b = 27b
✗
✗
1, 14
−1, 2
(b − 1) (14b + 2)
2b − 14b = −12b ✗
✗
1, 14
−2, 1
(b − 2) (14b + 1)
b − 28b = −27b ✗
✗
2, 7
1, −2
−4b + 7b = 3b
✓
✗
(2b + 1) (7b − 2)
2, 7
2, −1
(2b + 2) (7b − 1)
−2b + 14b = 12b
✗
2, 7
−1, 2
(2b − 1) (7b + 2)
2, 7
−2, 1
(2b − 2) (7b + 1)
✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗
4b − 7b = −3b
✗
2b − 14b = −12b ✗
(2b + 1)(7b − 2) = 0 2b + 1 = 0 −1 −1
or
7b − 2 = 0 +2 +2
2b = −1 2b −1 —=— 2 2 1 b = −— 2
7b = 2 7b 2 —=— 7 7 2 b=— 7 1 2 The roots are b = −— and b = —. 2 7
✗ ✓ ✗ ✗
Algebra 1 Worked-Out Solutions
421
Chapter 7 29. Let y = 0.
30. Let y = 0.
y = 2x2 − 3x − 35
y = 4x 2 + 11x − 3
0=
0 = 4x 2 + 11x − 3
2x2
− 3x − 35
Factors Factors Possible of 2 of −35 factorization
Factors Factors Possible of 4 of −3 factorization
Middle term
1, 2
1, −35
(x + 1) (2x − 35)
−35x + 2x = −33x ✗
1, 2
35, −1
(x + 35) (2x − 1)
−x + 70x = 69x
1, 2
−1, 35
(x − 1) (2x + 35)
35x − 2x = 33x
1, 2
−35, 1
(x − 35) (2x + 1)
1, 2
5, −7
(x + 5) (2x − 7)
−7x + 10x = 3x
1, 2
7, −5
(x + 7) (2x − 5)
−5x + 14x = 9x
1, 2
−5, 7
(x − 5) (2x + 7)
7x − 10x = −3x
✓
1, 2
−7, 5
(x − 7) (2x + 5)
5x − 14x = −9x
✗
1, 4
1, −3
(x + 1) (4x − 3)
−3x + 4x = x
✗
✗
1, 4
3, −1
(x + 3) (4x − 1)
−x + 12x = 11x
✓
✗
1, 4
−1, 3
(x − 1) (4x + 3)
3x − 4x = −x
✗
x − 70x = −69x ✗
1, 4
−3, 1
(x − 3) (4x + 1)
x − 12x = −11x
✗
✗
2, 2
1, −3
(2x + 1) (2x − 3)
−6x + 2x = −4x
✗
✗
2, 2
−1, 3
(2x − 1) (2x + 3)
6x − 2x = 4x
✗
0 = (x − 5)(2x + 7) x−5= 0 +5 +5
or
2x + 7 = 0 −7 −7
x=5
2x = −7 2x −7 —=— 2 2 7 x = −— 2 The x-coordinates of the points where the graph crosses the 7 x-axis are the roots x = 5 and x = −—. 2
422
Algebra 1 Worked-Out Solutions
Middle term
0 = (x + 3)(4x − 1) x+3= −3
0 −3
or
4x − 1 = +1
0 +1
x = −3
4x = 1 4x 1 —=— 4 4 1 x=— 4 The x-coordinates of the points where the graph crosses the 1 x-axis are the roots x = −3 and x = —. 4
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Chapter 7 31. Let y = 0.
32. Let y = 0.
y = −7x 2 − 2x + 5 0=
−7x 2
y = −3x 2 + 14x + 5
− 2x + 5
0 = −3x 2 + 14x + 5
0 = −(7x 2 + 2x − 5) −1(0) = −1[
−(7x 2
0 = −(3x 2 − 14x − 5)
+ 2x − 5) ]
−1(0) = −1[ −(3x 2 − 14x − 5) ]
0 = 7x 2 + 2x − 5
0 = 3x 2 − 14x − 5
Factors Factors Possible of 7 of −5 factorization
Factors Factors Possible of 3 of −5 factorization
Middle term
1, 7
1, −5
(x + 1) (7x − 5)
−5x + 7x = 2x
1, 7
5, −1
(x + 5) (7x − 1)
−x + 35x = 34x
1, 7
−1, 5
(x − 1) (7x + 5)
1, 7
−5, 1
(x − 5) (7x + 1)
✓
1, −5
(x + 1) (3x − 5)
−5x + 3x = −2x
✗
✗
1, 3
5, −1
(x + 5) (3x − 1)
−x + 15x = 14x
✗
5x − 7x = −2x
✗
1, 3
−1, 5
(x − 1) (3x + 5)
5x − 3x = 2x
✗
x − 35x = −34x
✗
1, 3
−5, 1
(x − 5) (3x + 1)
x − 15x = −14x
✓
0 = (x + 1)(7x − 5) x+1= −1
0 −1
or
Middle term
1, 3
0 = (x − 5)(3x + 1) 7x − 5 = +5
x = −1
0 +5
7x = 5 7x 5 —=— 7 7 5 x=— 7 The x-coordinates of the points where the graph crosses the 5 x-axis are the roots x = −1 and x = —. 7
x−5= +5
0
or
+5
3x + 1 = −1
0 −1
x=5
3x = −1 3x −1 —=— 3 3 1 x = −— 3 The x-coordinates of the points where the graph crosses the 1 x-axis are the roots x = 5 and x = −—. 3 33. a. 15x 2 − x − 2 = (3x + 1)(?)
Factors Factors of 15 of −2 Factorization 3, 5
1, −2
(3x + 1) (5x − 2)
Middle term −6x + 5x = −x
✓
So, 15x 2 − x − 2 = (3x + 1)(5x − 2) and the length of the sign is represented by (5x − 2) feet. b. One way to find the area of the sign when x = 3 is to
substitute 3 into the expression for the area 15x2 − x − 2 and then simplify. Another way is to substitute 3 into the expressions for the length (5x − 2) and width (3x + 1). Then simplify each expression and multiply these two values.
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Algebra 1 Worked-Out Solutions
423
Chapter 7 34. Let h = 0.
Factors of −1085
h = −16t2 + 8t + 80 0 = −16t2 + 8t + 80
w − 31 = + 31
Middle term
(t + 1) (2t − 10)
−10t + 2t = −8t
✗
1, 2
10, −1
(t + 10) (2t − 1)
−t + 20t = 19t
✗
1, 2
−1, 10
(t − 1) (2t + 10)
10t − 2t = 8t
✗
1, 2
−10, 1
(t − 10) (2t + 1)
1, 2
2, −5
(t + 2) (2t − 5)
1, 2
5, −2
(t + 5) (2t − 2)
1, 2
−2, 5
(t − 2) (2t + 5)
1, 2
−5, 2
(t − 5) (2t + 2)
t − 20t = −19t −5t + 4t = −t
✗
or
7, −155
−7, 155
−31, 35
31, −35
−1084
1084
−212
212
−148
148
4
−4
w + 35 =
0 + 31
− 35
w = 31
w = −35
36. Let w be the width of the invitation, then the length is 2w − 1.
A = ℓw 15 = (2w − 1)w
✗
15 − 15 = 2w 2 − w − 15
−2t + 10t = 8t
0 − 35
A negative width does not make sense, so you should use the positive solution. So, the width of the base is 31 meters, and the length is 8 + 2(31) = 70 meters.
✓
15 = 2w 2 − w 0 = 2w 2 − w − 15
5t − 4t = t 2t − 10t = −8t
2t − 5 = 0 +5 +5
t = −2
−5, 217
15 = 2w(w) − 1(w)
0 = (t + 2)(2t − 5) t+2= 0 −2 −2
5, −217
0 = (w − 31)(w + 35)
Factors Factors Possible of 2 of −10 factorization 1, −10
−1, 1085
Sum of factors
0 = −8(2t2 − t − 10) 1 1 −— (0) = −— [ −8(2t2 − t − 10) ] 8 8 0 = 2t2 − t − 10
1, 2
1, −1085
2t = 5 2t 5 —=— 2 2 t = 2.5
A negative time does not make sense, so you should use the positive solution. So, the diver is in the air for 2.5 seconds.
✗ ✗
Factors Factors Possible of 2 of −15 factorization 1, 2
1, −15
(w + 1) (2w − 15)
1, 2
15, −1
(w + 15) (2w − 1)
−w + 30w = 29w
✗
1, 2
−1, 15
(w − 1) (2w + 15)
15w − 2w = 13w
✗
1, 2
−15, 1
(w − 15) (2w + 1)
1, 2
3, −5
(w + 3) (2w − 5)
−5w + 6w = w
✗
1, 2
5, −3
(w + 5) (2w − 3)
−3w + 30w = 27w
✗
1, 2
−3, 5
(w − 3) (2w + 5)
5w − 6w = −w
✓
1, 2
−5, 3
(w − 5) (2w + 3)
3w − 10w = −7w
✗
35. Let w be the width of the base, then the length is 8 + 2w.
A = ℓw 2170 = (8 + 2w)(w) 2170 = 8(w) + 2w(w)
Middle term −15w + 2w = −13w ✗
w − 30w = −29w ✗
2170 = 8w + 2w2 2170 − 2170 = 8w + 2w2 − 2170 0 = 2w2 + 8w − 2170 0 = 2(w2 + 4w − 1085) 1 —2 (0)
= —12 [ 2(w2 + 4w − 1085) ]
0 = w2 + 4w − 1085
424
Algebra 1 Worked-Out Solutions
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Chapter 7 0 = (w − 3) (2w + 5) w−3= +3
0 +3
or
41. 2x 2 + tx + 10
2w + 5 = −5
0 −5
w=3
2w = −5 2w −5 —=— 2 2 5 w = −— 2 A negative width does not make sense, so you should use the positive solution. So, the width of the invitation is 3 inches, and the length is 2(3) − 1 = 6 − 1 = 5 inches. The invitation will fit inside the envelope because the width is less than 1 5 3 — inches, and the length is less than 5 — inches. 8 8 37. Sample answer: 3x(x + 2) = 3x(x) + 3x(2)
=
3x2
+ 6x
So, one binomial that has a GCF of 3x is 3x2 + 6x. 38. The graph of g represents function k, and the graph of h represents function ℓ. Because c is positive in each function,
the constant terms in the factors must have the same sign. Because g has a positive value of b, the constant terms of the factors will both be positive, which results in negative roots, and the graph of k has two negative x-intercepts. Because h has a negative value of b, the constant terms of the factors will both be negative, which results in positive roots, and the graph of ℓ has two positive x-intercepts.
Factors Factors Possible of 2 of 10 factorization
Middle term
1, 2
1, 10
(x + 1) (2x + 10)
10x + 2x = 12x
1, 2
−1, −10
(x − 1) (2x − 10)
−10x − 2x = −12x
1, 2
10, 1
(x + 10) (2x + 1)
x + 20x = 21x
1, 2
−10, −1
(x − 10) (2x − 1)
−x − 20x = −21x
1, 2
2, 5
(x + 2) (2x + 5)
5x + 4x = 9x
1, 2
−2, −5
(x − 2) (2x − 5)
−5x − 4x = −9x
1, 2
5, 2
(x + 5) (2x + 2)
2x + 10x = 12x
1, 2
−5, −2
(x − 5) (2x − 2)
−2x − 10x = −12x
So, the values of t that make 2x 2 + tx + 10 factorable are ±9, ±12, and ±21. 42 a.
39. It is not possible to factor ax 2 + bx + c, where a ≠ 1,
when no combination of factors of a and c produce the correct middle term. Sample answer: One such trinomial is 5x 2 − x + 3. 40. Your friend is incorrect. To use the Zero-Product Property,
2x 2 + 5x − 3 = (x + 3)(2x − 1) b.
one side of the equation needs to be 0. So, you must first subtract 2 from each side of the equation, then factor. 3x 2 − 2x − 1 = (x − 1)(3x + 1)
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Algebra 1 Worked-Out Solutions
425
Chapter 7 43. Let w be the width (in inches) of the rectangle, then the
44. Let x be the width (in feet) of the patio.
length is 2w + 1.
Area of patio
Area of = pool surface
A=5+P ℓw = 5 + 2ℓ + 2w
Area of Area of pool Area of pool = − pool surface surface and patio surface
(2w + 1)w = 5 + 2(2w + 1) + 2w 2w(w) + 1(w) = 5 + 2(2w) + 2(1) + 2w
(16)(24) = (x + 24 + x)(x + 16 + x) − (16)(24)
2w 2 + w = 5 + 4w + 2 + 2w 2w 2
384 = (2x + 24)(2x + 16) − 384
+ w = 4w + 2w + 5 + 2
384 = 2x(2x) + 2x(16) + 24(2x) + 24(16) − 384
2w 2 + w = 6w + 7 + w − 6w = 6w − 6w + 7
384 = 4x 2 + 32x + 48x + 384 − 384
2w 2 − 5w = 7
384 = 4x 2 + 80x
2w 2 − 5w − 7 = 7 − 7
384 − 384 = 4x 2 + 80x − 384
2w 2
0 = 4x 2 + 80x − 384
2w 2 − 5w − 7 = 0 Factors Factors Possible of 2 of −7 factorization
0 = 4( x 2 + 20x − 96 ) 1
1, 2
1, −7
(w + 1) (2w − 7)
−7w + 2w = −5w
✓
1, 2
7, −1
(w + 7) (2w − 1)
−w + 14w = 13w
✗
1, 2
−1, 7
(w − 1) (2w + 7)
7w − 2w = 5w
✗
1, 2
−7, 1
(w − 7) (2w + 1)
w − 14w = −13w ✗
−1
0 −1
w = −1
or
2w − 7 = +7
0 +7
2w = 7 2w 7 —=— 2 2 w = 3.5
A negative width does not make sense, so you should use the positive solution. So, the width of the rectangle is 3.5 inches.
[
]
0 = x 2 + 20x − 96 Factors of −96
Sum of factors
(w + 1)(2w − 7) = 0 w+1=
1
—4 (0) = —4 4( x 2 + 20x − 96 )
Middle term
1, −96
−1, 96
2, −48
−2, 48
3, −32
−3, 32
4, −24
−4, 24
6, −16
−6, 16
8, −12
−8, 12
−95
95
−46
46
−29
29
−20
20
−10
10
−4
4
0 = (x − 4)(x + 24) x−4= +4 x=
0 +4 4
or
x + 24 = − 24
0 − 24
x = −24
A negative width does not make sense, so you should use the positive solution. So, the width of the patio is 4 feet. 45. 4k 2 + 7jk − 2j 2
Factors Factors Possible of 4 of −2 factorization
Middle term
1, 4
1, −2
(k + j) (4k − 2j)
−2jk + 4jk = 2jk
✗
1, 4
2, −1
(k + 2j) (4k − j)
−jk + 8jk = 7jk
✓
1, 4
−1, 2
(k − j) (4k + 2j)
2jk − 4jk = −2jk
✗
1, 4
−2, 1
(k − 2j) (4k + j)
jk − 8jk = −7jk
✗
2, 2
1, −2
(2k + j) (2k − 2j)
−4jk + 2jk = −2jk
✗
2, 2
−1, 2
(2k − j) (2k + 2j)
4jk − 2jk = 2jk
✗
So, 4k 2 + 7jk − 2j 2 = (k + 2j)(4k − j).
426
Algebra 1 Worked-Out Solutions
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Chapter 7 46. 6x 2 + 5xy − 4y 2
48. 18m 3 + 39m 2n − 15mn 2 = 3m( 6m 2 + 13mn − 5n 2 )
Factors Factors Possible of 6 of −4 factorization
Factors Factors Possible of 6 of −5 factorization
Middle term
Middle term
1, 6
1, −4
(x + y) (6x − 4y)
−4xy + 6xy = 2xy
✗
1, 6
1, −5
(m + n) (6m − 5n)
−5mn + 6mn = mn
✗
1, 6
4, −1
(x + 4y) (6x − y)
−xy + 24xy = 23xy
✗
1, 6
5, −1
(m + 5n) (6m − n)
−mn + 30mn = 29mn
✗
1, 6
−1, 4
(x − y) (6x + 4y)
4xy − 6xy = −2xy
✗
1, 6
−1, 5
(m − n) (6m + 5n)
5mn − 6mn = −mn
✗
1, 6
−4, 1
(x − 4y) (6x + y)
xy − 24xy = −23xy ✗
1, 6
−5, 1
(m − 5n) (6m + n)
mn − 30mn = −29mn ✗
1, 6
2, −2
(x + 2y) (6x − 2y)
✗
2, 3
1, −5
(2m + n) (3m − 5n)
−10mn + 3mn = −7mn
✗
1, 6
−2, 2
(x − 2y) (6x + 2y)
2xy − 12xy = −10xy ✗
2, 3
5, −1
(2m + 5n) (3m − n)
−2mn + 15mn = 13mn
✓
2, 3
1, −4
(2x + y) (3x − 4y)
✗
2, 3
−1, 5
(2m − n) (3m + 5n)
10mn − 3mn = 7mn
✗
2, 3
4, −1
(2x + 4y) (3x − y)
✗
2, 3
−5, 1
(2m − 5n) (3m + n)
2mn − 15mn = −13mn ✗
2, 3
−1, 4
(2x − y) (3x + 4y)
2, 3
−4, 1
(2x − 4y) (3x + y)
2, 3 2, 3
2, −2
(2x + 2y) (3x − 2y)
−2, 2
(2x − 2y) (3x + 2y)
−2xy + 12xy = 10xy
−8xy + 3xy = −5xy −2xy + 12xy = 10xy 8xy − 3xy = 5xy
✓
2xy − 12xy = −10xy ✗ −4xy + 6xy = 2xy 4xy − 6xy = −2xy
✗ ✗
47. −6a 2 + 19ab − 14b 2 = − ( 6a 2 − 19ab + 14b 2 )
Middle term
−1, −14
(a − b) (6a − 14b)
−14ab − 6ab = −20ab ✗
1, 6
−14, −1
(a − 14b) (6a − b)
−ab − 84ab = −85ab ✗
1, 6
−2, −7
(a − 2b) (6a − 7b)
−7ab − 12ab = −19ab ✓
1, 6
−7, −2
(a − 7b) (6a − 2b)
−2ab − 42ab = −44ab ✗
2, 3
−1, −14
(2a − b) (3a − 14b)
−28ab − 3ab = −31ab ✗
2, 3
−14, −1
(2a − 14b) (3a − b)
−2ab − 42ab = −44ab ✗
2, 3
−2, −7
(2a − 2b) (3a − 7b)
−14ab − 6ab = −20ab ✗
−7, −2
(2a − 7b) (3a − 2b)
⋅
—
49. ± √ 64 = ± √ 8 8 = ± 8 —
⋅
—
50. √ 4 = √ 2 2 = 2
⋅ 52. ± √ 81 = ± √ 9⋅9 = ± 9 —
—
51. −√ 225 = −√ 15 15 = −15 —
—
53. y = 3 + 7x
y − x = −3 Step 1 Equation 1 is already solved for y. Step 2
y − x = −3 (3 + 7x) − x = −3
1, 6
2, 3
Maintaining Mathematical Proficiency —
So, 6x 2 + 5xy − 4y 2 = (2x − y)(3x + 4y).
Factors Factors Possible of 6 of 14 factorization
So, 18m 3 + 39m 2n − 15mn 2 = 3m (2m + 5n)(3m − n).
−4ab − 21ab = −25ab ✗
So, −6a 2 + 19ab − 14b 2 = −(a − 2b)(6a − 7b).
7x − x + 3 = −3 6x + 3 = −3 −3
−3
6x = −6 6x −6 —=— 6 6 x = −1 Step 3 y = 3 + 7x y = 3 + 7(−1) y=3−7 y = −4 Check
y = 3 + 7x ? −4 = 3 + 7(−1) ? −4 = 3 − 7
−4 = −4 ✓ The solution is (−1, −4). Copyright © Big Ideas Learning, LLC All rights reserved.
y − x = −3 ? −4 − (−1) = −3 ? −4 + 1 = −3 −3 = −3 ✓
Algebra 1 Worked-Out Solutions
427
Chapter 7 56. −x − 8 = −y
54. 2x = y + 2
9y − 12 + 3x = 0
−x + 3y = 14
Step 1
2x = y + 2
Step 1
2x − 2 = y + 2 − 2 2x − 2 = y −x + 3y = 14
Step 2
Step 2
−x + 3(2x − 2) = 14
9y − 12 + 3x = 0 9(x + 8)− 12 + 3x = 0
−x + 3(2x) − 3(2) = 14
9(x) + 9(8) − 12 + 3x = 0
−x + 6x − 6 = 14
9x + 72 − 12 + 3x = 0
5x − 6 = 14 +6
−x − 8 = −y −x − 8 −y —=— −1 −1 x+8=y
12x + 60 =
+6
− 60
5x = 20 5x 20 —=— 5 5 x=4 Step 3
12x = −60 12x −60 —=— 2 12 x = −5
2x = y + 2
Step 3
2(4) = y + 2
5 − 8 = −y
−2
−3 = −y −3 −y —=— −1 −1 3=y
6=y Check
−x − 8 = −y −(−5) − 8 = −y
8=y+2 −2
2x = y + 2 ? 2(4) = 6 + 2
−x + 3y = 14 ? −4 + 3(6) = 14 ? −4 + 18 = 14
8=8✓
14 = 14 ✓ The solution is (4, 6).
Check
−x − 8 = −y ? −(−5) − 8 = −3 ? 5 − 8 = −3 −3 = −3 ✓
55. 5x − 2y = 14
9y − 12 + 3x = 0 ? 9(3) − 12 + 3(−5) = 0 ? 27 − 12 − 15 = 0 ? 15 − 15 = 0 0=0✓
−7 = −2x + y Step 1
0 − 60
−7 = −2x + y
The solution is (−5, 3).
−7 + 2x = −2x + 2x + y 2x − 7 = y 5x − 2y = 14
Step 2
5x − 2(2x − 7) = 14 5x − 2(2x) − 2(−7) = 14 5x − 4x + 14 = 14 x + 14 = − 14 Step 3 −7 = −2x + y
14 − 14
x=0
−7 = −2(0) + y −7 = 0 + y −7 = y Check
5x − 2y = 14 ? 5(0) − 2(−7) = 14 ? 0 + 14 = 14 14 = 14 ✓
−7 = −2x + y ? −7 = −2(0) + (−7) ? −7 = 0 − 7 −7 = −7 ✓
The solution is (0, −7).
428
Algebra 1 Worked-Out Solutions
Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 7.7 Explorations (p. 397)
2.
1. a.
4x 2 − 1 = 4x 2 − 2x + 2x − 1 = (2x + 1)(2x − 1) This product is a sum and difference pattern. b.
9x 2 − 6x + 6x − 4 = 9x 2 − 4 = (3x + 2)(3x − 2)
4x 2 − 4x + 1 = (2x − 1)(2x − 1) = (2x − 1)2 This product is a square of a binomial pattern. c.
9x 2 − 6x − 6x + 4 = 9x 2 − 12x + 4 = (3x − 2)2
4x 2 + 4x + 1 = (2x + 1)(2x + 1) = (2x + 1)2 This product is a square of a binomial pattern. d.
9x 2 + 6x + 6x + 4 = 9x 2 + 12x + 4 = (3x + 2)2 3. Sample answer: The algebra tiles for special patterns will
4x 2 − 6x + 2 = (2x − 2)(2x − 1) This product is not a special pattern.
Copyright © Big Ideas Learning, LLC All rights reserved.
always form a square array, and the x 2 tiles will also form a square array. Factor special products by using the special patterns studied in Lesson 7.3, but in reversed order: For the sum and difference pattern, the number of x and −x tiles will be the same, and there will only be −1 tiles to complete the array. For the square of a binomial pattern, the x tiles will either be all positive or all negative and there will only be +1 tiles to complete the array.
Algebra 1 Worked-Out Solutions
429
Chapter 7 4. a.
+
+
+
+
+
+ +
+ +
+ +
+
2. 100 − m 2 = 10 2 − m 2
= (10 + m)(10 − m) So, 100 − m 2 = (10 + m)(10 − m).
+
3. 9n 2 − 16 = (3n)2 − 42
+
+
+
+
+
+
+
= (3n + 4)(3n − 4) So, 9n 2 − 16 = (3n + 4)(3n − 4).
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
5. 362 − 342 = (36 + 34)(36 − 34)
+
+
+
+
+
+
+
= 70(2)
4. 16h 2 − 49 = (4h)2 − 72
= (4h + 7)(4h − 7) So, 16h 2 − 49 = (4h + 7)(4h − 7).
= 140
25x 2 + 10x + 1 = (5x + 1)2 b.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
− 342 = 140.
6. 472 − 442 = (47 + 44)(47 − 44)
= 91(3) = 273 So, 472 − 442 = 273.
−
+
+
− −
+
So,
362
7. 552 − 502 = (55 + 50)(55 − 50)
+
+
+
+
+
+
−
+
+
+
+
+
+
−
+
+
+
+
+
+
−
−
−
−
−
−
−
+
= 105(5) = 525 So,
+
+
+
= 52(4) = 208 So, 282 − 242 = 208.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
− 502 = 525.
8. 282 − 242 = (28 + 24)(28 − 24)
25x 2 − 10x + 1 = (5x − 1)2 c.
552
+
+
9. m 2 − 2m + 1 = m 2 − 2(m)(1) + 12
= (m − 1)2 So,
m2
− 2m + 1 = (m − 1)2.
10. d 2 − 10d + 25 = d 2 − 2(d)(5) + 52
= (d − 5)2 So,
+
d2
− 10d + 25 = (d − 5)2.
11. 9z 2 + 36z + 36 = 9( z 2 + 4z + 4 )
+
+
+
+
+
+
= 9[ z 2 + 2(z)(2) + 22 ]
+
= 9(z + 2)2 +
+
+
+
+
+
+
−
−
−
−
−
−
−
( 25x 2 − 1 ) = (5x − 1)(5x + 1) 7.7 Monitoring Progress (pp. 398–400) 1.
x2
− 36 =
x2
−
62
= (x + 6)(x − 6) So, x 2 − 36 = (x + 6)(x − 6).
430
Algebra 1 Worked-Out Solutions
So, 9z 2 + 36z + 36 = 9(z + 2)2. a 2 + 6a + 9 = 0
12.
a2
+ 2(a)(3) + 32 = 0 (a + 3)2 = 0 a+3= −3
0 −3
a = −3 The equation has a repeated root of a = −3. Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 49 7 w 2 − —w + — = 0 3 36
13.
(
Monitoring Progress and Modeling with Mathematics 3. m 2 − 49 = m 2 − 72
) ( )
49 7 36 w 2 − —w + — = 36(0) 3 36 7 49 2 36( w ) − 36 —w + 36 — = 0 3 36
( )
= (m + 7)(m − 7) So,
= (z + 9)(z − 9)
(6w)2 − 2(6w)(7) + 72 = 0 (6w −
+7
So,
=0
6w − 7 =
14.
z2
− 81 = (z + 9)(z − 9).
5. 64 − 81d 2 = 82 − (9d)2
= (8 + 9d )(8 − 9d )
0
So, 64 − 81d 2 = (8 + 9d )(8 − 9d ).
+7
6w = 7 7 6w —=— 6 6 7 w=— 6 7 The equation has a repeated root of w = — . 6 n2
6. 25 − 4x 2 = 52 − (2x)2
= (5 + 2x)(5 − 2x) So, 25 − 4x 2 = (5 + 2x)(5 − 2x). 7. 225a 2 − 36b 2 = 9( 25a 2 − 4b 2 )
= 9[ (5a)2 − (2b)2 ]
− 81 = 0
= 9(5a + 2b)(5a − 2b)
n 2 − 92 = 0
So, 225a 2 − 36b 2 = 9(5a + 2b)(5a − 2b).
(n + 9)(n − 9) = 0 n+9= −9
0
n−9=
or
−9
+9
n = −9
0 +9
n=9
8. 16x 2 − 169y 2 = (4x)2 − (13y)2
= (4x + 13y)(4x − 13y) So, 16x 2 − 169y 2 = (4x + 13y)(4x − 13y). 9. 122 − 92 = (12 + 9)(12 − 9)
The roots are n = −9 and n = 9.
= 21(3)
15. Let y = 0.
= 63
y = 81 − 16t 2
So,
0 = 81 − 16t 2
122
= 30(8)
0 = (9 + 4t)(9 − 4t) 9 + 4t =
− 92 = 63.
10. 192 − 112 = (19 + 11)(19 − 11)
0 = 92 − (4t)2
−9
− 49 = (m + 7)(m − 7).
4. z 2 − 81 = z 2 − 92
36w 2 − 84w + 49 = 0
7)2
m2
0
9 − 4t =
or
−9
−9
4t = −9 4t −9 —=— 4 4 t = −2.25
= 240
0 −9
−4t = −9 −4t −9 —=— −4 −4 t = 2.25
A negative time does not make sense in this situation. So, the golf ball hits the ground after 2.25 seconds.
So, 192 − 112 = 240. 11. 782 − 722 = (78 + 72)(78 − 72)
= 150(6) = 900 So,
782
12. 542 − 522 = (54 + 52)(54 − 52)
= 106(2)
7.7 Exercises (pp. 401–402) Vocabulary and Core Concept Check 1. The square roots of the first and last terms are y and 8, and
⋅ ⋅
the middle term is 2 y 8, so it fits the perfect square trinomial pattern and can be factored as such.
= 212 So, 542 − 522 = 212. 13. 532 − 472 = (53 + 47)(53 − 47)
= 100(6)
2. The polynomial that does not belong is k 2 + 25, because it is
the only one that cannot be factored using a special pattern.
Copyright © Big Ideas Learning, LLC All rights reserved.
− 722 = 900.
= 600 So,
532
− 472 = 600.
Algebra 1 Worked-Out Solutions
431
Chapter 7 14. 392 − 362 = (39 + 36)(39 − 36)
25. a. d 2 + 8d + 16 = d 2 + 2(d)(4) + 42
= 75(3)
= (d + 4)2
= 225
So, an expression that represents the side length of the coaster is (d + 4) centimeters.
So, 392 − 362 = 225.
b. P = 4(s)
15. h 2 + 12h + 36 = h 2 + 2(h)(6) + 62
= 4(d + 4)
= (h + 6)2
= 4(d) + 4(4)
So, h 2 + 12h + 36 = (h + 6)2.
= 4d + 16
16. p 2 + 30p + 225 = p 2 + 2( p)(15) + 152
An expression for the perimeter of the coaster is (4d + 16) centimeters.
= ( p + 15)2 So, p 2 + 30p + 225 = ( p + 15)2.
26. a. A = x 2 − 30x + 225
= x 2 − 2(x)(15) + 152
17. y 2 − 22y + 121 = y 2 − 2(y)(11) + 112
= (x − 15)2
= ( y − 11)2
A polynomial that represents the side length of the playground is (x − 15) feet.
So, y 2 − 22y + 121 = (y − 11)2. 18. x 2 − 4x + 4 = x 2 − 2(x)(2) + 22
b. P = 4(s)
= (x − 2)2
= 4(x − 15)
So, x 2 − 4x + 4 = (x − 2)2.
= 4(x) − 4(15)
19. a 2 − 28a + 196 = a2 − 2(a)(14) + 142
= 4x − 60
= (a − 14)2
An expression for the perimeter of the playground is (4x − 60) feet.
So, a2 − 28a + 196 = (a − 14)2. 20. m 2 + 24m + 144 = m2 + 2(m)(12) + 122
27.
z 2 − 22 = 0
= (m + 12)2
(z + 2)(z − 2) = 0
So, m 2 + 24m + 144 = (m + 12)2.
z+2=
21. 25n 2 + 20n + 4 = (5n)2 + 2(5n)(2) + 22
= (5n +
z2 − 4 = 0
−2
2)2
23. The difference of two squares pattern should be used to
factor this polynomial. n 2 − 64 = n 2 − 82 = (n + 8)(n − 8) So,
n2
− 64 = (n + 8)(n − 8).
24. The perfect square trinomial pattern should be used to factor
this polynomial. y 2 − 6y + 9 = y 2 − 2(y)(3) + 32 = (y − 3)2 So,
432
y2
− 6y + 9 = ( y − 3)2.
Algebra 1 Worked-Out Solutions
0 +2
z=2
The roots are z = −2 and z = 2.
22. 49a 2 − 14a + 1 = (7a)2 − 2(7a)(1) + 12
So, 49a 2 − 14a + 1 = (7a − 1)2.
z−2= +2
z = −2
So, 25n 2 + 20n + 4 = (5n + 2)2. = (7a − 1)2
or
0 −2
4x 2 = 49
28.
4x 2
− 49 = 49 − 49
4x 2 − 49 = 0 (2x)2 − 72 = 0 (2x + 7)(2x − 7) = 0 2x + 7 = −7
0
or
−7
2x − 7 = +7
2x = −7 2x −7 —=— 2 2 7 x=−— 2 7 7 The roots are x = − — and x = —. 2 2
0 +7
2x = 7 2x 7 —=— 2 2 7 x=— 2
Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 29.
k 2 − 16k + 64 = 0
33.
k 2 − 2(k)(8) + 82 = 0 (k − 8)2 = 0 k−8= +8
0 +8
k=8 The equation has a repeated root of k = 8. 30.
s 2 + 20s + 100 = 0
1 1 y2 + —y = −— 16 2 1 1 1 1 y2 + —y + — = −— + — 16 16 2 16 1 1 y2 + —y + — = 0 2 16 1 1 16 y2 + —y + — = 16(0) 2 16 1 1 16( y2 ) + 16 —y + 16 — = 0 2 16 16y2 + 8y + 1 = 0
( ( )
(4y)2 + 2(4y)(1) + 12 = 0
s 2 + 2(s)(10) + 10 2 = 0
(4y + 1)2 = 0
(s + 10) 2 = 0 s + 10 = − 10
n2 + 9 = 6n n2 + 9 − 6n = 6n − 6n n2 − 6n + 9 = 0 n2 − 2(n)(3) + 32 = 0 (n − 3)2 = 0 n−3=
0 +3
n=3 The equation has a repeated root of n = 3. 32.
−1
− 10
The equation has a repeated root of s = −10.
+3
4y + 1 =
0
s = −10
31.
y 2 = 12y − 36 y 2 − 12y = 12y − 12y − 36 y 2 − 12y = −36
4y = −1
(
(3x − 2)2 = 0 3x − 2 = +2
(y − 6)2 = 0 +6
) ( ) ()
(3x)2 − 2(3x)(2) + 22 = 0
y 2 − 12y + 36 = 0 y 2 − 2(y)(6) + 62 = 0 0 +6
y=6 The equation has a repeated root of y = 6.
0 −1
4y −1 —=— 4 4 1 y = −— 4 1 The equation has a repeated root of y = −—. 4 4 4 34. −—x + — = −x2 3 9 4 4 −—x + — + x2 = −x2 + x2 3 9 4 4 x2 − —x + — = 0 3 9 4 4 9 x2 − —x + — = 9(0) 3 9 4 4 2 9( x ) − 9 —x + 9 — = 0 3 9 9x2 − 12x + 4 = 0
y 2 − 12y + 36 = −36 + 36
y−6=
) ( )
0 +2
3x = 2 3x 2 —=— 3 3 2 x=— 3 2 The equation has a repeated root of x = —. 3 35. 3z2 − 27 = 3( z2 − 9 ) = 3( z2 − 32 ) = 3(z + 3)(z − 3) So, 3z2 − 27 = 3(z + 3)(z − 3). 36. 2m2 − 50 = 2( m2 − 25 )
= 2( m2 − 52 ) = 2(m + 5)(m − 5) So,
Copyright © Big Ideas Learning, LLC All rights reserved.
2m2
− 50 = 2(m + 5)(m − 5).
Algebra 1 Worked-Out Solutions
433
Chapter 7 37. 4y 2 − 16y + 16 = 4( y 2 − 4y + 4 )
42. Let y = 1.
= 4[ y 2 − 2(y)(2) + 22 ]
y = −16t 2 + 8t
= 4( y −
1 = −16t 2 + 8t
2)2
So, 4y 2 − 16y + 16 = 4( y − 2)2.
1 + 16t 2 = −16t 2 + 16t + 8t
8k 2 + 80k + 200 = 8(k 2 + 10k + 25)
38.
= 8[ k + 2(k)(5) + 52 ] = 8(k + 5)2
39.
16t 2 − 8t + 1 = 0 (4t − 1)2 = 0
+ 120y + 72 = + 60y + 36) = 2[ (5y)2 + 2(5y)(6) + 62 ] 2(25y 2
= 2(5y + 6)2 So, 50y 2 + 120y + 72 = 2(5y + 6)2. 40. 27m2 − 36m + 12 = 3(9m2 − 12m + 4)
= 3[ (3m)2 − 2(3m)(2) + 22 ] = 3(3m − 2)2 So, 27m2 − 36m + 12 = 3(3m − 2)2. 41. Let y = 0.
4t − 1 = +1
Factors of 84
0 = (5 + 4t)(5 − 4t)
Sum of factors
0 −5
+1
4t = 1 4t 1 —=— 4 4 1 t = —, or 0.25 4 1 The equation has a repeated root of t = —. So, the 4 grasshopper is 1 foot off the ground 0.25 second after the start of the jump.
0 = 25 − 16t 2 5 + 4t =
0
43. a. w 2 + 18w + 84
y = 25 − 16t 2
−5
16t 2 + 1 − 8t = 8t − 8t (4t)2 − 2(4t)(1) + 12 = 0
So, 8k 2 + 80k + 200 = 8(k + 5)2. 50y 2
16t 2 + 1 = 8t
or
5 − 4t = −5
0 −5
4t = −5 −4t = −5 −4t −5 4t −5 —=— —=— 4 −4 −4 4 5 5 t = —, or 1.25 t = −—, or −1.25 4 4 A negative time does not make sense in this situation. So, the paintbrush lands on the ground 1.25 seconds after you drop it.
1, 84 2, 42 3, 28 4, 21 6, 14 7, 12 85
44
31
25
20
19
The trinomial w2 + 18w + 84 cannot be factored because no factor pair of 84 has a sum of 18. However, if you change the value of c to the perfect square 81, the trinomial w2 + 18w + 81 can be factored using the perfect square trinomial pattern. w2 + 18w + 81 = ( w2 + 2(w)(9) + 92 ) = (w + 9)2 So, w2 + 18w + 81 can be factored, but w2 + 18w + 84 cannot. b. y2 − 10y + 23
Factors of 23
−1, −23
Sum of factors
−24
The trinomial y 2 − 10y + 23 cannot be factored because the only factor pair of 23 does not have a sum of −10. However, if you change the value of c to the perfect square 25, the trinomial y2 − 10y + 25 can be factored using the perfect square trinomial pattern. y 2 − 10y + 25 = y 2 − 2(y)(5) + 52 = (y − 5)2 So, y 2 − 10y + 25 can be factored, but y 2 − 10y + 23 cannot.
434
Algebra 1 Worked-Out Solutions
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Chapter 7 44. a.
b. Set 9x2 − 144 equal to 81 and solve.
9x2 − 144 = 81 9x2
− 144 − 81 = 81 − 81 9x2 − 225 = 0 9( x2 − 25 ) = 0
—9 [ 9( x2 − 25 ) ] = —9 (0) 1
1
4x2 − 1 = (2x + 1)(2x − 1)
x2 − 25 = 0
b.
x2 − 52 = 0 (x + 5)(x − 5) = 0 x+5= 0 −5
+5
x = −5
4x2 − 4x + 1 = (2x − 1)2
0
+5
x=5
The side length cannot be negative. So, the solution is x = 5, which means that the side length of one picture frame is 5 inches.
45. To simplify (2x − 5)2 − (x − 4)2, you can square each
binomial and then combine like terms. Or, you can use the difference of two squares pattern with each binomial as one of the terms. Sample answer: Choose the difference of two squares pattern so that you do not need to square any binomials.
x−5=
or
−5
48. a.
Volume of the Volume Volume of other = + composite solid of cube rectangular prism = s 3 + ℓwh = (x)3 + 4(x)(4)
46. a. When you subtract b 2 from a 2, you “take away” the purple
= x 3 + 16x
region. So, the remaining blue, yellow, and red regions represent a 2 − b 2, as shown below.
A polynomial that represents the volume of the composite solid is ( x 3 + 16x ) cubic inches.
a+b
Yellow
Red
Blue
b. Set 25x equal to x 3 + 16x and solve.
25x = x 3 + 16x
a−b
25x − 25x = x 3 + 16x − 25x 0 = x 3 − 9x 0 = x( x 2 − 9 )
b. (a − b)2 is represented by the yellow region. a 2 is
0 = x( x 2 − 32 )
the entire square. Subtracting 2ab requires removing 2 regions that each represent ab, which would be the blue and purple regions, and the red and purple regions. There is only one purple region, and subtracting 2ab required removing it twice, so one purple region needs to be replaced, which adds b 2. 47. a. 9
⋅
0 = x(x + 3)(x − 3) x = 0 or x + 3 = −3
= 9( x2 − 16 ) =
9( x2 )
=
9x2
x−3= +3
x = −3
0
+3
x=3
Maintaining Mathematical Proficiency 49.
50
⋅ ⋅ ⋅
− 9(16)
10 5
− 144
A polynomial that represents the area of the picture frames, not including the pictures, is ( 9x2 − 144 ) square inches.
or
Side lengths cannot be negative or 0. So, the solution is x = 3.
Area of Area of frame Area of =9 − frame and picture picture = 9( x2 − 42 )
0 −3
2 5 5
⋅ ⋅
⋅
The prime factorization of 50 is 2 5 5, or 2 52. 50.
44
⋅ ⋅ ⋅
4 11 2 2 11
⋅ ⋅
⋅
The prime factorization of 44 is 2 2 11, or 22 11.
Copyright © Big Ideas Learning, LLC All rights reserved.
Algebra 1 Worked-Out Solutions
435
Chapter 7 51.
85
3y + 3 < x
56.
⋅
3y + 3 − 3 < x − 3
5 17 The prime factorization of 85 is 5 17.
⋅
52.
3y < x − 3 x−3 3 1 y < —x − 1 3 Now the inequality is in slope-intercept form. So, use the 1 1 slope m = — and the y-intercept b = −1 to graph y = — x − 1. 3 3 Because the symbol is <, use a dashed line, and shade below it. 3y 3
96
—<—
⋅ 12 2 ⋅4 ⋅ 4 ⋅3 2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅3 8
⋅ ⋅ ⋅ ⋅ ⋅
⋅
The prime factorization of 96 is 2 2 2 2 2 3, or 25 3. 53. Because the inequality is in slope-intercept form, use
the slope m = 4, and the y-intercept b = −1 to graph y = 4x − 1. Because the symbol is ≤ , use a solid line, and shade below it.
y −4
−2
y
4
−4
2
−4
4 x −2
−6
−2
2
4 x
7.8 Explorations (p. 403) 1. a. (x + 1)(x + 1)(−2) 54. Because the inequality is in slope-intercept form, use the slope m = − —12 and the y-intercept b = 3 to graph
y = − —12x + 3. Because the symbol is >, use a dashed line, and shade above it. 8
+
+
+
+
+
+
+
+
So, (x + 1)(x + 1)(−2) = ( x2 + 2x + 1 )(−2)
y
= x2(−2) + 2x(−2) + 1(−2)
6
= −2x2 − 4x − 2. b. (x + 2)(x + 1)(−x)
4
+
+
+
+
+
4y − 12 ≥ 8x
+ +
+ +
+
4y − 12 + 12 ≥ 8x + 12
2 x −2
55.
2
4
6
4y ≥ 8x + 12 4y 8x + 12 —≥— 4 4 y ≥ 2x + 3
+
So, (x + 2)(x + 1)(−x) = ( x2 + 3x + 2 )(−x) = x2(−x) + 3x(−x) + 2(−x) = −x3 − 3x2 − 2x.
Now the inequality is in slope-intercept form. So, use the slope m = 2 and the y-intercept b = 3 to graph y = 2x + 3. Because the symbol is ≥, use a solid line, and shade above it. 6
y
4
−4
2
4 x
−2
436
Algebra 1 Worked-Out Solutions
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Chapter 7 c. (x + 3)(x)(2)
2. a. J; x3 + x2 = x2(x + 1) b. A; x3 − x = x( x2 − 1 )
+ + + + +
= x( x2 − 12 ) = x(x + 1)(x − 1)
+ + + +
c. D;
x3
+
x2
d. G;
x3
−
4x2
= x(x + 2)(x − 1) + 4x = x( x2 − 4x + 4 ) = x( x2 − 2(x)(2) + 22 ) = x(x − 2)2
So, (x + 3)(x)(2) = ( x2 + 3x )(2) = x2(2) + 3x(2)
e. N; x3 − 2x2 − 3x = x( x2 − 2x − 3 )
= x(x + 1)(x − 3)
= 2x2 + 6x
f. B; x3 − 2x2 + x = x( x2 − 2x + 1 )
d. (x + 1)(x − 1)(x)
+
− 2x = x( x2 + x − 2 )
= x( x − 2(x)(1) + 1 )
−
+
+
−
+
+
−
= x(x − 1)2 g. F;
= x(x + 2)(x − 2) x2
+ (x − x) − 1 ](x)
= ( x2 − 1 )(x) = x2(x) − 1(x) = x3 − x. e. (−x + 1)(x + 1)(−x)
+
−
−
−
+
+
+
− 4x = x( x2 − 4 ) = x( x2 − 22 )
So, (x + 1)(x − 1)(x) = [
+
x3
h. L; x3 + 2x2 = x2(x + 2) i. I; x3 − x2 = x2(x − 1) j. E; x3 − 3x2 + 2x = x( x2 − 3x + 2 )
= x(x − 1)(x − 2) k. M; x3 + 2x2 − 3x = x( x2 + 2x − 3 )
= x(x + 3)(x − 1) l. O; x3 − 4x2 + 3x = x( x2 − 4x + 3 )
= x(x − 1)(x − 3) m. K; x3 − 2x2 = x2(x − 2) n. H; x3 + 4x2 + 4x = x( x2 + 4x + 4 )
So, (−x + 1)(x + 1)(−x) = [ −x2 + (x − x) + 1 ](−x) = ( −x2 + 1 )(−x) = −x2(−x) + 1(−x) = x3 − x. f. (−x − 1)(x + 1)(−2)
= x( x2 + 2(x)(2) + 22 ) = x(x + 2)2 o. C; x3 + 2x2 + x = x( x2 + 2x + 1 )
= x( x + 2(x)(1) + 12 ) = x(x + 1)2
+
Factor out the greatest common monomial factor first, then factor the remaining expression if possible.
−
−
−
3. Factor out the greatest common monomial factor first, then
−
−
−
+
factor the remaining expression if possible.
So, (−x − 1)(x + 1)(−2) = ( −x2 − 2x − 1 )(−2) = −x2(−2) − 2x(−2) − 1(−2) = 2x2+ 4x + 2.
Copyright © Big Ideas Learning, LLC All rights reserved.
Algebra 1 Worked-Out Solutions
437
Chapter 7 4. a. x3 + 4x2 + 3x = x( x2 + 4x + 3 )
= x(x + 1)(x + 3) So,
+
x( x2 − 25 ) = 0
+ 3x = x(x + 1)(x + 3). b. x3 − 6x2 + 9x = x( x2 − 6x + 9 ) x3
4x2
= x(
x2
− 2(x)(3) +
x3 − 25x = 0
7.
32
)
x( x2 − 52 ) = 0 x(x + 5)(x − 5) = 0 x=0
or
x+5=
= x(x − 3)2
−5
So, x3 − 6x2 + 9x = x(x − 3) 2. c. x3 + 6x2 + 9x = x( x2 + 6x + 9 ) = x( x2 + 2(x)(3) + 32 ) = x(x + 3)2 So, x3 + 6x2 + 9x = x(x + 3)2. 7.8 Monitoring Progress (pp. 404–406) 1.
a3
+
3a2
+a+3= =
( a3
+
a 2(a
+ 3) + (a + 3)
= (a + So,
+
a3
3a2
3)(a2
+ 1)
+ a + 3 = (a + 3)( a2 + 1 ).
2. y2 + 2x + yx + 2y = y2 + yx + 2x + 2y
= y( y + x) + 2(x + y) = y(x + y) + 2(x + y) = (x + y)( y + 2). So, y2 + 2x + yx + 2y = (x + y)( y + 2). 3. 3x3 − 12x = 3x( x2 − 4 )
= 3x(x + 2)(x − 2) So, 3x3 − 12x = 3x(x + 2)(x − 2). 4.
−
12y2
+ 18y = 2y(
y2
− 6y + 9 )
6.
−
− 8m = m(m − 4)(m + 2).
w( w2 − 8w + 16 ) = 0 w(w − 4)2 = 0 w−4= +4
0 +4
w=4 The equation has a root of w = 0 and a repeated root of w = 4.
438
c=0
or
c−3= +3
c−4=
or
0 +3
+4
c=3
0 +4
c=4
The roots are c = 0, c = 3, and c = 4. 9. Volume = length
72 =
x
⋅ ⋅
⋅ ⋅
width (x − 1)
72 = x(x − 1)(x + 9)
height (x + 9)
72 = x[ x(x) + x(9) − 1(x) − 1(9) ] 72 = x( x2 + 9x − x − 9 ) 72 = x( x2 + 8x − 9 ) 72 = x(x)2 + x(8x) − x(9) 72 − 72 = x3 + 8x2 − 9x − 72 0 = x3 + 8x2 − 9x − 72 0 = ( x3 + 8x2 ) + (−9x − 72) 0 = x2(x + 8) − 9(x + 8)
0 = (x + 8)(x + 3)(x − 3) x+8= −8
0 −8
x = −8
or
x+3= −3
0
or
−3
x = −3
x−3= +3
0 +3
x=3
Disregard x = −8 and x = −3 because x represents the length of the box, and a negative length does not make sense. So, 3 is the correct value of x. Use x = 3 to find the width and height, as shown.
w( w2 − 2(w)(4) + 42 ) = 0 or
c(c − 3)(c − 4) = 0
0 = (x + 8)( x2 − 32 )
w3 − 8w2 + 16w = 0
w=0
c( c2 − 7c + 12 ) = 0
= 2y( y −
= m(m − 4)(m + 2) So,
8. c3 − 7c2 + 12c = 0
0 = (x + 8)( x2 − 9 )
3)2
5. m3 − 2m2 − 8m = m( m2 − 2m − 8 )
2m2
x=5
= 2y( y − 2( y)(3) + 32 ) So, 2y3 − 12y2 + 18y = 2y( y − 3)2.
m3
x = −5
0 +5
72 = x3 + 8x2 − 9x
= 3x( x2 − 22 )
2y3
+5
The roots are x = 0, x = −5, and x = 5.
+ (a + 3)
3a2 )
x−5=
or
0 −5
Algebra 1 Worked-Out Solutions
width = x − 1
height = x + 9
=3−1
=3+9
=2
= 12
The length is 3 feet, the width is 2 feet, and the height is 12 feet.
Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 11. 2x3 − 2x = 2x( x2 − 1 )
7.8 Exercises (pp. 407–408)
= 2x( x2 − 12 )
Vocabulary and Core Concept Check
= 2x(x + 1)(x − 1)
1. It is written as a product of unfactorable polynomials with
So,
integer coefficients. 2. Look for terms with common factors.
2x3
− 2x = 2x(x + 1)(x − 1).
12. 36a 4 − 4a 2 = 4a 2( 9a 2 − 1 )
= 4a 2( (3a)2 − 12 )
Monitoring Progress and Modeling with Mathematics 3. x3 + x2 + 2x + 2 = ( x3 + x2 ) + (2x + 2)
= x2(x + 1) + 2(x + 1) = (x + 1)( x2 + 2 ) So,
x3
+
x2
+ 2x + 2 = (x + 1)( x2 + 2 ).
4. y3 − 9y2 + y − 9 = ( y3 − 9y2 ) + ( y − 9)
5.
−
So,
y3
9y2
3z3
+ 2z −
So, 36a 4 − 4a 2 = 4a 2(3a + 1)(3a − 1). 13. 2c2 − 7c + 19
Factors Factors of 2 of 19
Possible factorization
1, 2
−1, −19 (c − 1)(2c − 19) −19c − 2c = −21c ✗
= ( y − 9)(
1, 2
−19, −1 (c − 19)(2c − 1) −c − 38c = −39c ✗
y2
+ 1)
−8=
3z3
−
12z2
+ 2z − 8
= ( 3z3 − 12z2 ) + ( 2z − 8 ) = 3z2(z − 4) + 2(z − 4) = (z − 4)( 3z2 + 2 ) So, 3z3 + 2z − 12z2 − 8 = (z − 4)( 3z2 + 2 ). 6. 2s3 − 27 − 18s + 3s2 = 2s3 + 3s2 − 18s − 27
= ( 2s3 + 3s2 ) + (−18s − 27) = s2(2s + 3) − 9(2s + 3) = (2s + 3)( s2 − 9 )
The polynomial 2c2 − 7c + 19 is unfactorable because no combination of factors of 2 and 19 produce a middle term of −7c. 14. m2 − 5m − 35
Factors of −35
1, −35
−1, 35
5, −7
−5, 7
Sum of factors
−34
34
−2
2
The polynomial m2 − 5m − 35 is unfactorable because a = 1, and no factor pairs of −35 have a sum of −5. 15. 6g3 − 24g2 + 24g = 6g( g2 − 4g + 4 )
= 6g( g2 − 2(g)(2) + 2 2 )
= (2s + 3)( s2 − 32 )
= 6g(g − 2) 2
= (2s + 3)(s + 3)(s − 3) So, 2s3 − 27 − 18s + 3s2 = (2s + 3)(s + 3)(s − 3). 7. x2 + xy + 8x + 8y = ( x2 + xy ) + (8x + 8y)
So,
= (x + y)(x + 8) So,
+ xy + 8x + 8y = (x + y)(x + 8).
6g3
= (q + 1)(q + 5p) So,
q2
+ q + 5pq + 5p = (q + 1)(q + 5p).
9. m2 − 3m + mn − 3n = ( m2 − 3m ) + (mn − 3n)
= m(m − 3) + n(m − 3) = (m − 3)(m + n) So, m2 − 3m + mn − 3n = (m − 3)(m + n). 10. 2a 2 + 8ab − 3a − 12b = ( 2a 2 + 8ab ) + (−3a − 12b)
= 2a(a + 4b) − 3(a + 4b) = (a + 4b)(2a − 3) So, 2a 2 + 8ab − 3a − 12b = (a + 4b)(2a − 3). Copyright © Big Ideas Learning, LLC All rights reserved.
+ 24g = 6g(g − 2)2. = −3d(d − 1)(5d − 2)
So,
−15d 3
+
21d 2
− 6d = −3d(d − 1)(5d − 2).
17. 3r 5 + 3r 4 − 90r 3 = 3r 3( r 2 + r − 30 )
= 3r 3(r − 5)(r + 6)
8. q2 + q + 5pq + 5p = ( q2 + q ) + (5pq + 5p)
= q(q + 1) + 5p(q + 1)
−
24g2
16. −15d 3 + 21d 2 − 6d = −3d( 5d 2 − 7d + 2 )
= x(x + y) + 8(x + y) x2
Middle term
= y2( y − 9) + ( y − 9) + y − 9 = ( y − 9)( y2 + 1 ).
12z2
= 4a 2(3a + 1)(3a − 1)
So,
3r 5
+
3r 4
− 90r 3 = 3r 3(r − 5)(r + 6).
18. 5w4 − 40w3 + 80w2 = 5w2( w2 − 8w + 16 )
= 5w2( w2 − 2(w)(4) + 4 2 ) = 5w2(w − 4) 2 So, 5w4 − 40w3 + 80w2 = 5w2(w − 4) 2. 19. −4c4 + 8c3 − 28c2 = −4c2( c2 − 2c + 7 )
Factors of 7 Sum of factors
−1, −7 −8
c2 − 2c + 7 is unfactorable, so the polynomial is factored completely. So, −4c4 + 8c3 − 28c2 = −4c2( c2 − 2c + 7 ).
Algebra 1 Worked-Out Solutions
439
Chapter 7 20. 8t2 + 8t − 72 = 8( t2 + t − 9 )
Factors of −9
x3 + x2 = 4x + 4
25.
x3 + x2 − 4x = 4x − 4x + 4
−1, 9 1, −9 −3, 3
Sum of factors
8
x3 + x2 − 4x = 4
0
−8
x3 + x2 − 4x − 4 = 4 − 4
t2 + t − 9 is unfactorable, so the polynomial is factored completely. So, 8t2 + 8t − 72 = 8( t2 + t − 9 ).
x3 + x2 − 4x − 4 = 0
( x3 + x2 ) + (−4x − 4) = 0 x2(x + 1) − 4(x + 1) = 0 (x + 1)( x2 − 4 ) = 0
21. b3 − 5b2 − 4b + 20 = ( b3 − 5b2 ) + (−4b + 20)
=
b2(b
(x + 1)( x2 − 2 2 ) = 0
− 5) − 4(b − 5)
= (b − 5)( b2 − 4 )
(x + 1)(x + 2)(x − 2) = 0
= (b −
x+1=
5)( b2
−
22 )
−1
= (b − 5)(b + 2)(b − 2) So,
b3
−
5b2
= (h + 4)( h2 − 5 2 )
2t3(t + 9)(t − 8) = 0 2t3 = 0 or 2t3 0 —=— 2 2 t3 = 0
− 25h − 100 = (h + 4)(h + 5)(h − 5).
23. 5n3 − 30n2 + 40n = 0
5n( n2
3—
√t3
− 6n + 8 ) = 0 n−2=
0
+2 +2
+4
n=2
+4 27.
n=4
k 2(k + 10)(k − 10) = 0 0
√k 2 = √0 k=0
or
k + 10 = − 10
−9
0
or
−9
t−8=0 +8
t = −9
+8
t=8
= √0
12s − 3s3 = 0
3s(2 + 5)(2 − 5) = 0
k 2( k 2 − 10 2 ) = 0
—
3—
t+9=
3s( 2 2 − s2 ) = 0
k 2( k 2 − 100 ) = 0
k2 =
x=2
3s( 4 − s2 ) = 0
k 4 − 100k 2 = 0
—
+2
The equation has roots t = −9, t = 8, and a repeated root t = 0.
or n − 4 = 0
The roots are n = 0, n = 2, and n = 4. 24.
+2
t=0
5n(n − 2)(n − 4) = 0 5n = 0 or 5n 0 —=— 5 5 n=0
x−2= 0
x = −2
2t3( t2 + t − 72 ) = 0
4)( h2
or
−2
The roots are x = −1, x = −2, and x = 2.
= (h + 4)(h + 5)(h − 5) +
−2
− 25 )
= (h +
So,
x+2= 0
26. 2t5 + 2t4 − 144t3 = 0
= h2(h + 4) − 25(h + 4)
4h2
or
x = −1
− 4b + 20 = (b − 5)(b + 2)(b − 2).
22. h3 + 4h2 − 25h − 100 = ( h3 + 4h2 ) + (−25h − 100)
h3
0
−1
0 − 10
k = −10
or k − 10 =
0
+ 10 + 10
3s = 0 or 2 + s = 0 3s 0 −2 −2 —=— 3 3 s=0 s = −2
or
2−s= 0 +s
+s
2=s
The roots are s = 0, s = −2, and s = 2.
k = 10
The equation has roots of k = −10, k = 10, and repeated roots of k = 0.
440
Algebra 1 Worked-Out Solutions
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Chapter 7 4y3 − 7y2 + 28 = 16y
28.
31. Let y = 0.
4y3 − 7y2 + 28 − 16y = 16y − 16y −
y =−2x4 + 16x3 − 32x2
− 16y + 28 = 0
0 = −2x4 + 16x3 − 32x2
( 4y3 − 7y2 ) + (−16y + 28) = 0
0 = −2x2( x2 − 8x + 16 )
4y3
y2(4y
7y2
− 7) − 4(4y − 7) = 0
0 = −2x2( x2 − 2(x)(4) + 4 2 )
(4y − 7)( y2 − 4 ) = 0
0 = −2x2(x − 4) 2
(4y − 7)(
y2
−
22
)=0
(4y − 7)( y + 2)( y − 2) = 0 4y − 7 = +7
or y + 2 =
0 +7
−2
4y = 7 4y 7 —=— 4 4 7 y=— 4
0
or
−2
y−2= +2
y = −2
0 +2
y=2
= √0
x=0
0 = x( 4x2 + 25x − 56 ) 0 = x(x + 8)(4x − 7) x=0
− 81 )
or
x+8= −8
0 = x( x2 − 9 2 ) x+9= −9
0
x−9= 0
or
−9
+9
x = −9
+9
x=9
The x-coordinates of the points where the graph crosses the x-axis are the roots x = 0, x = −9, and x = 9. 30. Let y = 0.
+7
0 +7
4x = 7 4x 7 —=— 4 4 7 x=— 4 The x-coordinates of the points where the graph crosses the 7 x-axis are the roots x = 0, x = −8, and x = —. 4
= a 2(a + 8) − 6(a + 8)
0 = −3x2( x2 + 8x + 15 )
= (a + 8)( a 2 − 6 )
0 = −3x2(x + 3)(x + 5)
—
4x − 7 =
a 3 + 8a 2 − 6a − 48 = ( a 3 + 8a 2 ) + (−6a − 48)
0 = −3x4 − 24x3 − 45x2
—
−8
or
33. In the second group, factor out −6 instead of 6.
y = −3x4 − 24x3 − 45x2
or
0
x = −8
0 = x(x + 9)(x − 9)
−3x2 = 0 0 −3x2 —=— −3 −3 x2 = 0
x=4
—
0 = 4x3 + 25x2 − 56x
0 = x3 − 81x
or
+4
y = 4x3 + 25x2 − 56x
y = x3 − 81x
x=0
√ x2
+4
0
32. Let y = 0.
29. Let y = 0.
0=
—
x−4=
The x-coordinates of the points where the graph crosses the x-axis are the repeated roots x = 0 and x = 4.
7 The roots are y = —, y = −2, and y = 2. 4
x( x2
−2x2 = 0 or 0 −2x2 —=— −2 −2 x2 = 0
x+3= −3
0 −3
x = −3
or x + 5 = −5
0 −5
x = −5
√ x2 = √ 0
x=0 The x-coordinates of the points where the graph crosses the x-axis are the roots x = −5, x = −3, and the repeated root x = 0.
Copyright © Big Ideas Learning, LLC All rights reserved.
So,
a3
+
8a 2
− 6a − 48 = (a + 8)( a 2 − 6 ).
34. It is not factored completely because x2 − 9 can be factored.
x3 − 6x2 − 9x + 54 = ( x3 − 6x2 ) + (−9x + 54) = x2(x − 6) − 9(x − 6) = (x − 6)( x2 − 9 ) = (x − 6)( x2 − 3 2 ) = (x − 6)(x + 3)(x − 3) So, x3 − 6x2 − 9x + 54 = (x − 6)(x + 3)(x − 3).
Algebra 1 Worked-Out Solutions
441
Chapter 7 35. a. Volume = length
=
4
⋅ ⋅
⋅ height ⋅ (w + 4)
width w
= 4w(w + 4)
Volume =
36.
⋅ ⋅
⋅ ⋅
width w
1152 = w(2w + 4)(18 − w)
height (18 − w)
= 4w(w) + 4w(4)
1152 = w[ 2w(18) + 2w(−w) + 4(18) + 4(−w) ]
=
1152 = w( 36w − 2w2 + 72 − 4w )
+ 16w
4w2
1152 = w( −2w2 + 32w + 72 ) 1152 = w( −2w2 ) + w(32w) + w(72)
A polynomial that represents the volume of the birdhouse is ( 4w2 + 16w ) cubic inches. b.
length
1152 = (2w + 4)
128 = 4w2 + 16w − 128
1152 = −2w3 + 32w2 + 72w
128 − 128 = 4w2 + 16w − 128w 0=
4w2
1152 − 1152 = −2w3 + 32w2 + 72w − 1152
+ 16w − 128
0 = −2w3 + 32w2 + 72w − 1152
0 = 4( w2 + 4w − 32 ) 1 —4 (0)
=
0=
1 —4
0 = −2( w3 − 16w2 − 36w + 576 )
[ 4( w2 + 4w − 32 ) ]
−—12 (0)
+ 4w − 32 ( 0 = w − 4 )( w + 8 ) w2
w−4= +4
0 +4
or
= −—12 (−2)( w3 − 16w2 − 36w + 576 )
0 = ( w3 − 16w2 − 36w + 576 )
w+8= −8
w=4
0 = ( w3 − 16w2 ) + (−36w + 576) 0
0 = w2(w − 16) − 36(w − 16)
−8
0 = (w − 16)( w2 − 36 )
w = −8
Disregard w = −8 because a negative width does not make sense, so 4 is the correct value of w. height = w + 4 =4+4 =8 The length is 4 inches, the width is 4 inches, and the height is 8 inches.
0 = (w − 16)( w2 − 6 2 ) 0 = (w − 16)(w + 6)(w − 6) w − 16 = + 16
w+6=
or
0 + 16
−6
w = 16
0
or w − 6 =
−6
+6
w = −6
0 +6
w=6
Disregard w = −6 because a negative width does not make sense. You know the height is greater than the width. Test the solutions of the equation, 16 and 6, in the expression for height. height = 18 − w = 18 − 16 = 2 ✗ or height = 18 − w = 18 − 6 = 12 ✓ The solution 6 gives a height of 12 inches, which is greater than 6, so 6 is the correct value of w. length = 2w + 4 = 2(6) + 4 = 12 + 4 = 16 The width is 6 inches, the length is 16 inches, and the height is 12 inches. 37. x3 + 2x2y − x − 2y = ( x3 + 2x2y ) + (−x − 2y)
= x2(x + 2y) − (x + 2y) = (x + 2y)( x2 − 1 ) = (x + 2y)( x2 − 12 ) = (x + 2y)(x + 1)(x − 1) So,
x3
+
2x2y
− x − 2y = (x + 2y)(x + 1)(x − 1).
38. 8b3 − 4b2a − 18b + 9a = ( 8b3 − 4b2a ) + (−18b + 9a)
= 4b2(2b − a) − 9(2b − a) = (2b − a)( 4b2 − 9 ) = (2b − a)[ (2b) 2 − 3 2 ] So,
442
Algebra 1 Worked-Out Solutions
8b3
−
4b2a
= (2b − a)(2b + 3)(2b − 3) − 18b + 9a = (2b − a)(2b + 3)(2b − 3). Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 39. 4s2 − s + 12st − 3t = ( 4s2 − s ) + (12st − 3t)
46. a. width = 4 + h
= s(4s − 1) + 3t(4s − 1)
length = 9 − h
= (4s − 1)(s + 3t) So, 4s2 − s + 2st − 3t = (4s − 1)(s + 3t).
Volume =
length
=
(9 − h)
= ( m2 − 2n )(6m + n)
= h(36) + h(5h) − h( h2 )
41. no; x3 + 2x2 + 3x + 6 = 0
= 36h + 5h2 − h3
( x3 + 2x2 ) + (3x + 6) = 0
= −h3 + 5h2 + 36
x2(x + 2) + 3(x + 2) = 0
A polynomial that represents the volume of the box in terms of its height is ( −h3 + 5h2 + 36h ) cubic inches.
(x + 2)( x2 + 3 ) = 0 The factors of the polynomial are x + 2 and x2 + 3. Using the zero product property, x + 2 = 0 will give one real solution x = −2. However, x2 + 3 = 0 has no real solutions because when you subtract 3 from each side you get x2 = −3, and no real number multiplied by itself will give a negative number.
−
h3 − 5h2 − 36h + 180 = 0
( − 5h2 ) + (−36h + 180) = 0 h2(h − 5) − 36(h − 5) = 0
+5
0 = x(x − 3)(x + 3)(x − 2) −3
x=3
−3
+2
0
or
h+6=
+5
−6
h=5
0 or x − 2 =
x = −3
− 36h = −h3 + h3 + 5h2 − 5h2 + 36h − 36h
h3
h−5=
y = x(x − 3)(x + 3)(x − 2) 0 or x + 3 =
5h2
(h − 5)(h + 6)(h − 6) = 0
y = x4 − 2x3 − 9x2 + 18x
0 +2
x=2
The x-intercepts of the graph of the function are 0, 3, −3, and 2. 43. a. Sample answer:
x2(x + 1) + 1(x + 2) = x2(x) + x2(1) + 1(x) + 1(2) = x3 + x2 + x + 2 Because the original expression did not have a common binomial, x3 + x2 + x + 2 is not factorable.
0 −6
h = −6
or
h−6= +6
0 +6
h=6
Disregard h = −6 because a negative height does not make sense. So, the height is either 5 inches or 6 inches. If h = 5, the width = 4 + h = 4 + 5 = 9 and the length = 9 − h = 9 − 5 = 4. If h = 6, the width = 4 + h = 4 + 6 = 10 and the length = 9 − h = 9 − 6 = 3. So, one possibility is that the length is 4 inches, the width is 9 inches, and the height is 5 inches. Another possibility is that the length is 3 inches, the width is 10 inches, and the height is 6 inches. c. SA = 2ℓw + 2ℓh + 2wh
= 2(4)(9) + 2(4)(5) + 2(9)(5)
b. Sample answer:
x2(x
180 +
h3
(h − 5)( h2 − 62 ) = 0
equal to 0 and solve for x to get the x-coordinates of the x-intercepts.
+3
180 = −h3 + 5h2 + 39h
b.
(h − 5)( h2 − 36 ) = 0
42. The x-intercepts occur when y = 0, so set each factor
+3
h
= h( 36 + 5h − h2 )
So, 6m3 − 12mn + m2n − 2n2 = ( m2 − 2n )(6m + n).
x−3=
height
= h[ 9(4) + 9(h) − h(4) − h(h) ] = h( 36 + 9h − 4h − h2 )
= 6m( m2 − 2n ) + n( m2 − 2n )
or
⋅ ⋅
width (4 + h)
= h(9 − h)(4 + h)
40. 6m3 − 12mn + m2n − 2n2 = ( 6m3 − 12mn ) + ( m2n − 2n2 )
x=0
⋅ ⋅
+ 1) + 1(x + 1) =
x2(x)
+
x2(1)
+ 1(x) + 1(1)
= x3 + x2 + x + 1 Because the original expression had a common binomial, x3 + x2 + x + 1 can be factored by grouping. 44. Your friend is not correct because it is possible that the terms
have a common monomial factor. 45. 12z3 − 27z = 3z( 4z2 − 9 )
= 3z[ (2z) 2 − 3 2 ] = 3z(2z + 3)(2z − 3) Expressions that could represent the dimensions of the room are 3z, 2z + 3, and 2z − 3. Copyright © Big Ideas Learning, LLC All rights reserved.
= 72 + 40 + 90 = 202 SA = 2ℓw + 2ℓh + 2wh = 2(3)(10) + 2(3)(6) + 2(10)(6) = 60 + 36 + 120 = 216 The box that has a length of 4 inches, a width of 9 inches, and a height of 5 inches is the box with the least possible surface area. These dimensions result in a surface area of 202 square inches. The other possible dimensions result in a surface area of 216 square inches.
Algebra 1 Worked-Out Solutions
443
Chapter 7 V = π r 2h
47.
b. Sample answer: 5x3 + wx2 + 80x = 0
25hπ = π (h − 3)2h
w 5x x2 + — x + 16 = 0 5
(
25hπ = π (h − 3)2h
π
π
Factors of 16
25h = (h − 3)2h
Sum of factors
25h = [ h2 − 2(h)(3) + 32 ]h
⋅
25h = h2(h) − 6h(h) + 9h 25h = h3 − 6h2 + 9h 25h − 25h = h3 − 6h2 + 9h − 25h 0 = h( h2 − 6h − 16 ) 0 = h(h − 8)(h + 2) h−8= +8
2, 8
4, 4
17
10
8
⋅
The equation has three solutions if w = 50. When w = 50, factoring out 5x will leave a factorable trinomial that is not a perfect square trinomial, so there will be three factors.
0 = h3 − 6h2 − 16h
or
1, 16
w Let — = 10. 5 w 5 — = 5 10 5 w = 50
25h = ( h2 − 6h + 9 )h
h=0
)
0
or
+8
h=8
h+2= −2
0 −2
h = −2
Disregard h = 0 and h = −2 because the height cannot be zero or negative. So, the height is 8 units and the radius is 8 − 3 = 5 units. 48. x5 − x 4 − 5x 3 + 5x 2 + 4x − 4
Maintaining Mathematical Proficiency 50. Use m = 1 and b =−4 to graph y = x − 4.
Use m = −2 and b = 2 to graph y = −2x + 2. y
−4
−2
4 x
2
= ( x5 − x4 ) + ( −5x 3 + 5x 2 ) + (4x − 4) = x 4(x − 1) − 5x 2(x − 1) + 4(x − 1)
y = −2x + 2
2
(2, −2)
−2
y=x−4
= (x − 1)( x 4 − 5x 2 + 4 ) = (x − 1)( x 2 − 4 )( x 2 − 1 )
The solution is the point of intersection, (2, −2).
= (x − 1)( x 2 − 2 2 )( x 2 − 12 ) = (x − 1)(x − 2)(x + 2)(x − 1)(x + 1) So,
x5
−
x4
−
5x 3
+
5x 2
+ 4x − 4
1
Use m = 3 and b = −3 to graph y = 3x − 3.
= (x − 1)2(x + 1)(x − 2)(x + 2). 4
49. a. Sample answer: 5x 3 + wx 2 + 80x = 0
w 5x x 2 + — x + 16 = 0 5
(
)
w Let — = 2(1)(4). 5 w —=8 5 w 5 —=8 5 5 w = 40
⋅
1
51. Use m = —2 and b = 2 to graph y = —2 x + 2.
⋅
y
(2, 3) y = 1x + 2 2
−2
2
4
x
−2
y = 3x − 3
The solution is the point of intersection, (2, 3).
The equation has two solutions if w = 40. When w = 40, factoring out 5x will leave a perfect square trinomial, so there will be two factors.
444
Algebra 1 Worked-Out Solutions
Copyright © Big Ideas Learning, LLC All rights reserved.
Chapter 7 1 x
()
1
5x − y = 12
—4 x + y = 9
5x − 5x − y = 12 − 5x
—4 x − —4 x + y = 9 − —4 x
52.
1
1
55. y = 9 —3 1
y = −—14x + 9
−y = −5x + 12
−2
x
1 −2 3
x
( ) 9( — )
9 —13
−1(−y) = −1(−5x + 12)
y
y = 5x − 12
0
−1
()
9 —13
27
9
9 —13
81
Use m = 5 and b = −12 to graph y = 5x − 12. 1
−1
(4, 8)
1
8
4
4
−8
−4
8 x
4
The solution is the point of intersection, (4, 8). 53. x = 3y
y − 10 = 2x
x 3
y − 10 + 10 = 2x + 10
3y 3
1 3
—x = y
1 4 3
3
1
—3
1
—9
1
1 x
2
6 x
4
−2
−1
0
1
2
−3(0.5)−2
−3(0.5)−1
−3(0.5)0
−3(0.5)1
−3(0.5)2
y
−12
−6
−3
−1.5
−0.75
From the graph, you can see the domain is all real numbers and the range is y < 0.
y
y = 2x + 10
−2
6 x
4
y = −3(0.5)x −8 −12
y = 2x + 10
−16
y = 1x 3
4
57. f(x) = −3(4)x
−8
4 x
−2
x
−4
−1
0
1
2
−3(4)x −3(4)−2 −3(4)−1 −3(4)0 −3(4)1 −3(4)2
(−6, −2)
The solution is the point of intersection, (−6, −2).
f(x)
3
3
−— 16
54. f (x) = 5x
−—4
−3
−12
−48
y −6
x
−2
−1
0
1
2
5x
5−2
5−1
50
51
52
f (x)
1 — 25
1 —5
1
5
25
16
From the graph, you can see the domain is all real numbers and the range is y > 0.
y
12 8 4
−2
1 3 3
−3(0.5)x
y
−4
1 2 3
x
1 1 Use m = — and b = 0 to graph y = — x. 3 3 Use m = 2 and b = 10 to graph y = 2x + 10.
−12
1 1 3
56. y = −3(0.5)x
y = 5x − 12
—=—
4
y = 9( 3)
−2
−4
3
From the graph, you can see the domain is all real numbers and the range is y > 0.
y
y = −4 x + 9
2
( ) 9( — ) 9( — ) 9( — ) 9( — )
y
1
Use m = −—4 and b = 9 to graph y = −—4 x + 9.
0
1
−4
−2
2 x −8
f(x) = −3(4)x
−16 −24 −32
From the graph, you can see the domain is all real numbers and the range is y < 0. 7.5 –7.8 What Did You Learn? (p. 409)
f(x) = 5x 2
4 x
Copyright © Big Ideas Learning, LLC All rights reserved.
1. The solutions of the equation m2 + 3m + 2 = 0 are the
m-coordinates of the m-intercepts of the graph of y = m2 + 3m + 2.
Algebra 1 Worked-Out Solutions
445
Chapter 7 2. The solutions of the equation 280 = −x2 + 38x are
x = 10 and x = 28. However, the diagram shows that the width of the road is less than 18 meters. So, a width of 28 meters does not make sense in this situation. Therefore, the only reasonable solution is x = 10. So, the width of the road is 10 meters.
10.
F
= 3y 2 − 7y − 40 11. (x + 4)( x2 + 7x ) = x( x2 ) + x(7x) + 4( x2 ) + 4(7x)
= x3 + 7x2 + 4x2 + 28x = x3 + 11x2 + 28x 12. (−3y + 1)( 4y 2 − y − 7 )
The greatest degree is 2. So, the degree of the polynomial is 2.
4y 2 − y − 7
The leading coefficient is 2.
×
The polynomial has 2 terms. So, it is a binomial.
−3y + 1 4y 2
2. You can write the polynomial −3p3 + 5p6 − 4 in standard
−y−7
−12y 3 + 3y 2 + 21y
form as 5p6 − 3p3 − 4.
−12y 3 + 7y 2 + 20y − 7
The greatest degree is 6. So, the degree of the polynomial is 6. The leading coefficient is 5.
13. (x + 9)(x − 9) = x2 − 92
= x2 − 81
The polynomial has 3 terms. So, it is a trinomial. 3. You can write the polynomial 9x7 − 6x2 + 13x5 in standard
form as
L
= 3y 2 + 8y + (−15y) + (−40)
1. You can write the polynomial 6 + 2x2 in standard form as
9x7
I
(y − 5)(3y + 8) = y(3y) + y(8) + (−5)(3y) + (−5)(8)
Chapter 7 Review (pp. 410–412) 2x2 + 6.
O
+
13x5
−
14. (2y + 4)(2y − 4) = (2y)2 − 42
= 4y 2 − 16
6x2.
The greatest degree is 7. So, the degree of the polynomial is 7. The leading coefficient is 9.
15. ( p + 4)2 = p 2 + 2( p)(4) + 42
= p 2 + 8p + 16
The polynomial has 3 terms. So, it is a trinomial. 4. You can write the polynomial −12y + 8y3 in standard form
16. (−1 + 2d )2 = (−1)2 + 2(−1)(2d) + (2d )2
= 1 − 4d + 4d 2
as 8y3 − 12y.
= 4d 2 − 4d + 1
The greatest degree is 3. So, the degree of the polynomial is 3.
17. x 2 + 5x = 0
The leading coefficient is 8.
x(x + 5) = 0
The polynomial has 2 terms. So, it is a binomial.
x=0
5. (3a + 7) + (a − 1) = 3a + a + 7 − 1
or
x+5=0 −5
= (3a + a) + (7 − 1) = 4a + 6 6. (x2 + 6x − 5) + (2x2 + 15) = x2 + 2x2 + 6x − 5 + 15
= (x2 + 2x2) + 6x + (−5 + 15)
The roots are x = 0 and x = −5. 18. (z + 3)(z − 7) = 0
z+3=0
= 3x2 + 6x + 10
−3
7. (−y2 + y + 2) − (y2 − 5y − 2)
= −2y2 + 6y + 4 8. ( p + 7) − (6p 2 + 13p) = p + 7 − 6p 2 − 13p
= −6p 2 + ( p − 13p) + 7 = −6p 2 − 12p + 7 9. (x + 6)(x − 4) = x(x − 4) + 6(x − 4)
or
−3
z−7=0 +7
z = −3
= −y2 + y + 2 − y2 + 5y + 2 = (−y2 − y2) + (y + 5y) + (2 + 2)
−5
x = −5
+7
z=7
The roots are z = −3 and z = 7. (b + 13)2 = 0
19.
(b + 13)(b + 13) = 0 b + 13 = 0 −13
−13
b = −13
or
b + 13 = 0 −13
−13
b = −13
The equation has repeated roots b = −13.
= x(x) + x(−4) + 6(x) + 6(−4) = x2 + (−4x) + 6x + (−24) = x2 + 2x − 24
446
Algebra 1 Worked-Out Solutions
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Chapter 7 20. 2y( y − 9)( y + 4) = 0
2y = 0 2y 0 —=— 2 2 y=0
or
26. −5y 2 − 22y − 8 = −(5y 2 + 22y + 8)
y−9=0
y+4=0
or
+9 +9
−4
y=9
Factors Factors of 5 of 8
−4
y = −4
The roots are y = 0, y = 9, and y = −4. 21. p 2 + 2p − 35
Factors of −35
–1, 35
1, –35
–5, 7
5, –7
Sum of factors
34
–34
2
–2
So,
b2
1, 80
2, 40
4, 20
5, 16
8, 10
81
42
24
21
18
+ 18b + 80 = (b + 8)(b + 10).
Factors of −21
–1, 21
1, –21
–3, 7
3, –7
Sum of factors
20
–20
4
−4
So, 24.
x2
1, 8
(y + 1)(5y + 8)
8y + 5y = 13y
✗
1, 5
8, 1
(y + 8)(5y + 1)
y + 40y = 41y
✗
1, 5
2, 4
(y + 2)(5y + 4)
4y + 10y = 14y
✗
1, 5
4, 2
(y + 4)(5y + 2)
2y + 20y = 22y
✓
Factors Factors of 6 of 7
23. z 2 − 4z − 21
z2
1, 5
27. 6x 2 + 17x + 7
22. b 2 + 18b + 80
Sum of factors
–1, –28
–2, –14
–4, –7
–29
–16
–11
25. 3t 2 + 16t − 12
Possible factorization
1, 6
1, 7
(x + 1)(6x + 7)
7x + 6x = 13x
✗
1, 6
7, 1
(x + 7)(6x + 1)
x + 42x = 43x
✗
2, 3
1, 7
(2x + 1)(3x + 7)
14x + 3x = 17x
✓
2, 3
7, 1
(2x + 7)(3x + 1)
2x + 21x = 23x
✗
Factors Factors of 2 of 6
So, x2 − 11x + 28 = (x − 4)(x − 7).
Factors Factors of 3 of −12
Middle term
28. −2y2 + 7y − 6 = −(2y2 − 7y + 6)
− 11x + 28
Sum of factors
Possible factorization
So, 6x 2 + 17x + 7 = (2x + 1)(3x + 7).
− 4z − 21 = (z + 3)(z − 7).
Factors of 28
Middle term
So, −5y2 − 22y − 8 = −(y + 4)(5y + 2).
So, p 2 + 2p − 35 = (p − 5)(p + 7).
Factors of 80
Possible factorization
Possible factorization
Middle term
1, 2
–1, –6
(y – 1)(2y – 6)
–6y – 2y = –8y
✗
1, 2
–6, –1
(y – 6)(2y – 1)
–y – 12y = –13y
✗
1, 2
–2, –3
(y – 2)(2y – 3)
–3y – 4y = –7y
✓
1, 2
–3, –2
(y – 3)(2y – 2)
–2y – 6y = –8y
✗
So, −2y 2 + 7y − 6 = −(y − 2)(2y − 3). 29. 3z 2 + 26z − 9
Middle term
1, 3
1, –12
(t + 1)(3t – 12)
–12t + 3t = –9t
✗
1, 3
12, –1
(t + 12)(3t – 1)
–t + 36t = 35t
✗
1, 3
–1, 12
(t – 1)(3t + 12)
12t – 3t = 9t
✗
1, 3
1, 3
–12, 1
(t – 12)(3t + 1)
✗
1, 3
2, –6
(t + 2)(3t – 6)
–6t + 6t = 0
1, 3
6, –2
(t + 6)(3t – 2)
1, 3
–2, 6
(t – 2)(3t + 6)
1, 3
–6, 2
(t – 6)(3t + 2)
1, 3
3, –4
(t + 3)(3t – 4)
–4t + 9t = 5t
✗
1, 3
4, –3
(t + 4)(3t – 3)
–3t + 12t = 9t
✗
1, 3
–3, 4
(t – 3)(3t + 4)
4t – 9t = –5t
✗
1, 3
–4, 3
(t – 4)(3t + 3)
9t – 12t = –3t
✗
Possible factorization
Middle term
1, –9
(z + 1)(3z – 9)
–9z + 3z = –6z
✗
1, 3
9, –1
(z + 9)(3z – 1)
–z + 27z = 26z
✓
✗
1, 3
–1, 9
(z – 1)(3z + 9)
9z – 3z = 6z
✗
–2t + 18t = 16t
✓
1, 3
–9, 1
(z – 9)(3z + 1)
z – 27z = –26z
✗
6t – 6t = 0
✗
1, 3
3, –3
(z + 3)(3z – 3)
✗
1, 3
–3, 3
(z – 3)(3z + 3)
t – 36t = –35t
2t – 18t = –16t
Factors Factors of 3 of −9
–3z + 9z = 6z 3z – 9z = –6z
✗ ✗
So, 3z 2 + 26z − 9 = (z + 9)(3z − 1).
So, 3t 2 + 16t − 12 = (t + 6)(3t − 2). Copyright © Big Ideas Learning, LLC All rights reserved.
Algebra 1 Worked-Out Solutions
447
Chapter 7 30. 10a 2 − 13a − 3
38. 3x 3 − 9x 2 − 54x = 0
Factors Factors of 10 of −3
3x(x 2 − 3x − 18) = 0
Possible factorization
3x(x − 6)(x + 3) = 0
Middle term
1, 10
1, –3
(a + 1)(10a – 3)
–3a + 10a = 7a ✗
1, 10
3, –1
(a + 3)(10a – 1)
–a + 30a = 29a ✗
1, 10
–1, 3
(a – 1)(10a + 3)
3a – 10a = –7a
✗
3x = 0 3x 0 —=— 3 3 x=0
1, 10
–3, 1
(a – 3)(10a + 1)
a – 30a = –29a
✗
The roots are x = 0, x = 6, and x = −3.
2, 5
1, –3
(2a + 1)(5a – 3)
–6a + 5a = –a
✗
2, 5
3, –1
(2a + 3)(5a – 1) –2a + 15a = 13a ✗
2, 5
–1, 3
(2a – 1)(5a + 3)
2, 5
–3, 1
(2a – 3)(5a + 1)
2a – 15a = –13a ✓
−3
−3
x = −3
or
2x − 3 = 0
−3
+3
+3
2x = −3 2x −3 —=— 2 2 3 x = −— 2
2x = 3 2x 3 —=— 2 2 3 x=— 2 3 3 The roots are x = −— and x = —. 2 2
y 2 − 100 = y 2 − 102 = (y + 10)(y − 10) So, y 2 − 100 = (y + 10)(y − 10). z 2 − 6z + 9 = z2 − 2(z)(3) + 32 = (z − 3)2
40.
So, z 2 − 6z + 9 = (z − 3)2.
z 3 + 3z 2 − 25z − 75 = 0 (z 3 + 3z 2) + (−25z − 75) = 0 z 2(z + 3) − 25(z + 3) = 0
m 2 + 16m + 64 = m 2 + 2(m)(8) + 82
(z + 3)(z 2 − 25) = 0
= (m + 8)2
(z + 3)(z 2 − 52) = 0
So, m 2 + 6m + 64 = (m + 8)2.
(z + 3)(z + 5)(z − 5) = 0
n 3 − 9n = n(n 2 − 9)
35.
x=6
2x + 3 = 0
So, x 2 − 9 = (x + 3)(x − 3).
34.
−3
(2x + 3)(2x − 3) = 0
= (x + 3)(x − 3)
33.
+6
(2x)2 − 32 = 0
x 2 − 9 = x 2 − 32
32.
+6
x+3=0
or
4(4x 2 − 9) = 0 1 1 — [ 4(4x 2 − 9) ] = —(0) 4 4 4x 2 − 9 = 0
So, 10a 2 − 13a − 3 = (2a − 3)(5a + 1). 31.
x−6=0
16x 2 − 36 = 0
39.
✗
6a – 5a = a
or
z+3=0
= n(n 2 − 32)
−3
= n(n + 3)(n − 3)
−3
z = −3
So, n 3 − 9n = n(n + 3)(n − 3).
or
z+5=0 −5
or
−5
z = −5
z−5=0 +5 +5 z=5
The roots are z = −3, z = −5, and z = 5.
x 2 − 3x + 4ax − 12a = (x 2 − 3x) + (4ax − 12a)
36.
= x(x − 3) + 4a(x − 3) = (x − 3)(x + 4a) So, x 2 − 3x + 4ax − 12a = (x − 3)(x + 4a). 37. 2x 4 + 2x 3 − 20x 2 = 2x 2(x 2 + x − 10)
Factors of −10
–1, 10
1, –10
–2, 5
2, –5
Sum of factors
9
–9
3
–3
+ x − 10 is unfactorable, so the original polynomial is factored completely. So, 2x4 + 2x3 − 20x2 = 2x2(x2 + x − 10).
x2
448
Algebra 1 Worked-Out Solutions
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Chapter 7
⋅ width ⋅ height 96 = (x + 8) ⋅ x ⋅ (x − 2)
5.
41. Volume = length
F
= 6w 2 + w − 15 So, (2w − 3)(3w + 5) = 6w 2 + w − 15.
96 = x[ x 2 + (−2x + 8x) − 16 ] 96 = x(x 2 + 6x − 16)
96 − 96 =
+
+ x(6x) + x(−16)
= z 2 − 121
6x 2
So, (z + 11)(z − 11) = z 2 − 121.
− 16x − 96
7. Check the first and last terms to verify they are perfect
0 = (x 3 + 6x 2) + (−16x − 96)
squares. If they are, find 2 times the product of their square roots. If this matches the middle term, or the opposite of the middle term, it is a perfect square trinomial.
0 = x 2(x + 6) − 16(x + 6) 0 = (x + 6)(x 2 − 16) 0 = (x + 6)(x 2 − 42)
8. A monomial is a type of polynomial, and a number is a type
0 = (x + 6)(x + 4)(x − 4) x+6=0 −6
(z + 11)(z − 11) = z 2 − 112
6.
96 = x 3 + 6x 2 − 16x x3
x+4=0
or
−6
−4
x = −6
of monomial. So, 18 is a polynomial. or
x−4=0
−4
+4
x = −4
+4
9. s 2 − 15s + 50
x=4
Disregard x = −6 and x = −4 because a negative width does not make sense. So, the solution is x = 4, which means the length of the box is 4 + 8 = 12 feet, the width is 4 feet, and the height is 4 − 2 = 2 feet.
Factors of 50 Sum of factors
–1, –50 –2, –25 –5, –10 –51
–27
–15
So, s 2 − 15s + 50 = (s − 5)(s − 10). h 3 + 2h 2 − 9h − 18 = (h 3 + 2h 2) + (−9h − 18)
10.
= h 2(h + 2) − 9(h + 2)
Chapter 7 Test (p. 413)
= (h + 2)(h 2 − 9)
1. (−2p + 4) − (p 2 − 6p + 8) = −2p + 4 − p 2 + 6p − 8
= (h + 2)(h 2 − 32)
= −p2 + (−2p + 6p) + (4 − 8) = −p 2 + 4p − 4 The greatest degree is 2. So, the degree of the polynomial is 2. The polynomial has 3 terms. So, it is a trinomial. 2. (9c6 − 5b 4) − (4c6 − 5b 4) = 9c6 − 5b 4 − 4c6 + 5b 4
= (9c6 − 4c6) + (−5b 4 + 5b 4)
= (h + 2)(h + 3)(h − 3) So, h 3 + 2h 2 − 9h − 18 = (h + 2)(h + 3)(h − 3). 11. −5k 2 − 22k + 15 = −(5k 2 + 22k − 15)
Factors Factors of 5 of −15
Possible factorization
Middle term
= 5c6 + 0
1, 5
1, –15
(k + 1)(5k – 15) –15k + 15k = 10k
✗
=
1, 5
15, –1
(k + 15)(5k – 1)
–k + 75k = 74k
✗
1, 5
–1, 15
(k – 1)(5k + 15)
15k – 5k = 10k
✗
1, 5
–15, 1
(k – 15)(5k + 1)
1, 5
3, –5
(k + 3)(5k – 5)
–5k + 15k = 10k
✗
= 6s 4 + 0 + (−3t)
1, 5
5, –3
(k + 5)(5k – 3)
–3k + 25k = 22k
✓
= 6s 4 − 3t
1, 5
–3, 5
(k – 3)(5k + 5)
5k – 15k = –10k ✗
The greatest degree is 4. So, the degree of the polynomial is 4.
1, 5
–5, 3
(k – 5)(5k + 3)
3k – 25k = –22k ✗
The polynomial has 2 terms. So, it is a binomial.
So, −5k 2 − 22k + 15 = −(k + 5)(5k − 3).
5c6
The expression has a degree of 6. The polynomial has 1 term. So, it is a monomial. 3. (4s 4 + 2st + t) + (2s 4 − 2st − 4t)
= (4s 4 + 2s 4) + (2st − 2st) + (t − 4t)
4.
L
= 6w 2 + (10w − 9w) + (−15)
96 = x[x(x) + x(−2) + 8(x) + 8(−2)]
96 =
I
(2w − 3)(3w + 5) = 2w(3w) + 2w(5) + (−3)(3w) + (−3)(5)
96 = x(x + 8)(x − 2)
x(x 2)
O
k – 75k = –74k ✗
(h − 5)(h − 8) = h(h − 8) − 5(h − 8) = h(h) + h(−8) + (−5)(h) + (−5)(−8) = h2 + (−8h − 5h) + 40 = h2 − 13h + 40 So, (h − 5)(h − 8) = h2 − 13h + 40.
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449
Chapter 7 12. (n − 1)(n + 6)(n + 5) = 0
n−1=0 +1
16. a. x 2 + 27x + 176
n+6=0
or
+1
−6
n=1
n+5=0
or
−6
−5
n = −6
−5
n = −5
d 2 + 2(d)(7) + 72 = 0
c.
−7
=
Sum of factors
177
90
48
30
27
⋅
Area = length width 1 10 = 2x x + — 2 1 10 = 2x x + — 2 1 10 = 2x(x) + 2x — 2 10 = 2x 2 + x
(
The equation has repeated roots of d = −7. +
11, 16
⋅(
d = −7
14.
8, 22
2x + 32 + 2x + 22 = (4x + 54) feet.
d+7=0
8x 2
4, 44
b. The perimeter is P = 2(x + 16) + 2(x + 11) =
(d + 7)2 = 0
6x 4
2, 88
Because the length of the stage is represented by the binomial (x + 16) feet, a binomial for the width of the stage is (x + 11) feet.
d 2 + 14d + 49 = 0
−7
1, 176
So, x 2 + 27x + 176 = (x + 11)(x + 16).
The roots are n = 1, n = −6, and n = −5. 13.
Factors of 176
26x 3
6x 4 + 8x 2 − 26x 3 = 26x 3 − 26x 3
)
)
()
6x 4 − 26x 3 + 8x 2 = 0
10 − 10 = 2x 2 + x − 10
2x2(3x 2 − 13x + 4) = 0
0 = 2x 2 + x − 10
Factors Factors of 3 of 4 –1, –4
1, 3
Possible factorization (x – 1)(3x – 4)
Factors Factors Possible of 2 of −10 factorization
Middle term –4x – 3x = –7x
✗
1, 3
–4, –1
(x – 4)(3x – 1)
–x – 12x = –13x
✓
1, 3
–2, –2
(x – 2)(3x – 2)
–2x – 6x = –8x
✗
2x 2(x − 4)(3x − 1) = 0 2x 2 = 0 2x 2 0 —=— 2 2 x2 = 0 —
or
x−4=0 +4
or
+4
x=4 —
√x2 = √0 x=0
3x − 1 = 0 +1
1, 2
1, –10
(x + 1) (2x – 10)
–10x + 2x = –8x
1, 2
10, –1
(x + 10) (2x – 1)
–x + 20x = 19x
1, 2
–1, 10
(x – 1) (2x + 10)
10x – 2x = 8x
1, 2
–10, 1
(x – 10) (2x + 1)
1, 2
2, –5
(x + 2) (2x – 5)
–5x + 4x = –x
1, 2
5, –2
(x + 5) (2x – 2)
–2x + 10x = 8x
1, 2
–2, 5
(x – 2) (2x + 5)
5x – 4x = x
1, 2
–5, 2
(x – 5) (2x + 2)
2x – 10x = –8x
+1
3x = 1 3x 1 —=— 3 3 1 x=— 3
1 The equation has roots of x = 4, x = —, and repeated roots 3 of x = 0. 15. π(r − 3)2 = π(r − 3)(r − 3)
= π[ r(r) + r(−3) + (−3)(r) + (−3)(−3) ] = π[ r 2 + (−3r − 3r) + 9 ] = π (r 2 − 6r + 9) = π (r 2) − π (6r) + π (9) = πr 2 − 6πr + 9π So, a polynomial that represents the area covered by the hour hand of the clock in one rotation is (πr 2 − 6πr + 9π) square units.
Middle term ✗ ✗ ✗ ✗
x – 20x = –19x
✗ ✗ ✓ ✗
0 = (x − 2)(2x + 5) x–2=0 +2
+2
or
2x + 5 = 0 –5
–5
x=2
2x = –5 –5 2x — =— 2 2 5 x=–— 2 5 5 Disregard x = −— because 2x = 2 −— = −5 feet does 2 2 not make sense as the length of the trap door. So, the solution is x = 2.
( )
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Chapter 7 d. Area of stage = x 2 + 27x + 176
= (2)2 + 27(2) + 176 = 4 + 54 + 176 = 234 ft2
(
1 Area of trap door = 2x x + — 2
( ) ()
20 Area of trapdoor = 20(10) = 200
ft2
The area of the stage is 234 square feet, and 20 times the area of the trap door is 200 square feet. So, the magician’s requirement is satisfied because the area of the stage is greater than 20 times the area of the trap door. 17. Sample answer: A polynomial equation in factored form that
has three positive roots is (x − 1)(x − 2)(x − 3) = 0. 18. Let y = 0.
( x 3 + 3x 2 ) + (−16x − 48) = 0 x 2(x + 3) − 16(x + 3) = 0 (x + 3)( x 2 − 16 ) = 0 (x + 3)( x 2 − 42 ) = 0
(x + 3)(x + 4)(x − 4) = 0 x+3=0 −3
−3
x = −3
or
x+4=0 −4
or
x−4=0
−4
+4
x = −4
+4
x=4
Disregard x = −3 and x = −4 because widths of −3 − 2 = −5 and −4 − 2 = −6, respectively, do not make sense. So, the solution is x = 4. length = x + 6 =4+6 = 10 width = x − 2 =2
0 = −16t 2 + 24t
height = x − 1
0 = −8t(2t − 3)
=4−1 2t − 3 = 0
or
x 3 + 3x 2 − 16x − 48 = 0
=4−2
y = −16t 2 + 24t
−8t = 0 0 −8t —=— −8 −8 t=0
x 3 + 3x 2 − 16x + 12 − 60 = 60 − 60
)
1 = 2(2) 2 + — 2 5 = 2(2) — 2 = 10 ft2
⋅
x 3 + 3x2 − 16x + 12 = 60
b.
+3
+3
2t = 3 2t 3 —=— 2 2 t = 1.5
You leave the surface of the trampoline when t = 0, then you are in the air, and then you make contact with the surface of the trampoline again after 1.5 seconds. So, you are in the air for 1.5 − 0 = 1.5 seconds.
=3 So, the length of the box is 10 inches, the width is 2 inches, and the height is 3 inches. Chapter 7 Standards Assessment (pp. 414–415) 1. a. The polynomial −4x 3 has 1 term. So, it is a monomial. b. The polynomial 6y − 3y5 has 2 terms. So, it is a
binomial. c. The polynomial c2 + 2 + c has 3 terms. So, it is a
trinomial.
⋅ width ⋅ height = (x + 6) ⋅ (x − 2) ⋅ (x − 1)
d. The polynomial −10d 4 + 7d 2 has 2 terms. So, it is a
= (x + 6)[x(x) + x(−1) + (−2)(x) + (−2)(−1)]
f. The polynomial 3b6 − 12b8 + 4b4 has 3 terms. So, it is a
19. a. Volume = length
= (x + 6)(x − 2)(x − 1)
= (x + 6)[ x 2 + (−x − 2x) + 2 ] = (x + 6)( x 2 − 3x + 2 ) = x(x2) + x(−3x) + x(2) + 6( x2 ) + 6(−3x) + 6(2) = x 3 − 3x 2 + 2x + 6x 2 − 18x + 12 = x 3 + ( −3x 2 + 6x2 ) + ( 2x − 18x ) + 12 =
x3
+
3x 2
− 16x + 12
So, a polynomial that represents the volume of the box is ( x 3 + 3x 2 − 16x + 12 ) cubic inches.
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binomial. e. The polynomial −5z11 + 8z12 has 2 terms. So, it is a
binomial. trinomial. The order of the polynomials by degree from least to greatest is c, a, d, b, f, and e. 2. D; Over the interval x = 0 to x = 2, the function f increases
by 25 − 4 = 21, the function g increases by 18 − 8 = 10, the function h increases by 9 − 1 = 8, and the function j increases by 72 − 2 = 70. So, the function j increases at the fastest rate over this interval.
Algebra 1 Worked-Out Solutions
451
Chapter 7 x 3 + 6x 2 − 4x = 24
3.
6. −2x + 15x 2 − 8 = 15x 2 − 2x − 8
x 3 + 6x 2 − 4x − 24 = 24 − 24
Factors Factors of 15 of −8
x 3 + 6x 2 − 4x − 24 = 0
Possible factorization
Middle term
( x3 + 6x2 ) + (−4x − 24) = 0
1, 15
1, −8
(x + 1)(15x − 8)
−8x + 15x = 7x
✗
x 2(x + 6) − 4(x + 6) = 0
1, 15
8, −1
(x + 8)(15x − 1)
−x + 120x = 119x
✗
1, 15
−1, 8
(x − 1)(15x + 8)
8x − 15x = −7x
✗
1, 15
−8, 1
(x − 8)(15x + 1)
x − 120x = −119x ✗
1, 15
2, −4
(x + 2)(15x − 4)
−4x + 30x = 26x
✗
1, 15
4, −2
(x + 4)(15x − 2)
−2x + 60x = 58x
✗
1, 15
−2, 4
(x − 2)(15x + 4)
4x − 30x = −26x
✗
1, 15
−4, 2
(x − 4)(15x + 2)
2x − 60x = −58x
✗
3, 5
1, −8
(3x + 1)(5x − 8)
−24x + 5x = −19x
✗
3, 5
8, −1
(3x + 8)(5x − 1)
−3x + 40x = 37x
✗
3, 5
8, −1
(3x − 1)(5x + 8)
24x + 5x = 19x
✗
3, 5
−8, 1
(3x − 8)(5x + 1)
3x − 40x = −37x
✗
3, 5
2, −4
(3x + 2)(5x − 4) −12x + 10x = −2x
✓
3, 5
4, −2
(3x + 4)(5x − 2)
−6x + 20x = 14x
✗
3, 5
−2, 4
(3x + 2)(5x + 4)
12x − 10x = 2x
✗
3, 5
−4, 2
(3x − 4)(5x + 2)
(x + 6)( x 2 − 4 ) = 0 (x + 6)( x 2 − 22 ) = 0 (x + 6)(x + 2)(x − 2) = 0 x+6=0 −6
or
x+2=0
−6
−2
x = −6
x−2=0
or
−2
+2
x = −2
+2
x=2
The solutions of the equation are x = −6, x = −2, and x = 2. +1
4.
+1
+1
+1
+1
Hours, x
1
2
3
4
5
6
Distance (miles), y
62
123
184
245
306
367
+61 +61 +61 +61 +61 As x increases by 1, y increases by 61. So there is a constant rate of change, which means the table can be represented by 61 a linear function with m = — = 61. 1 y − y1 = m( x − x1 ) y − 62 = 61(x − 1) y − 62 = 61(x) − 61(1)
= (3x + 2)(5x − 4) = (5x − 4)(3x + 2).
+ 62
7. a. The graph is going up from left to right. So, the function
y = 61x + 1
is increasing for increasing values of x.
So, an equation that models the distance traveled y as a function of the number of hours x is y = 61x + 1. 5. a. Use m =
1 −—3 4
and b = 2 to graph y =
1 −—3 x
b. From the graph, you can see that the range of the function
is y > 0. So, the graph does not have an x-intercept. The graph appears to cross the y-axis at (0, 2). To verify, check that f(0) equals 2: f(0) = 2(3)0 = 2(1) = 2. So, the y-intercept is 2.
+ 2.
y
8. B;
x 2 + 6x − 2 ×
−4
−2
✗
So, −2x + 15x 2 − 8 = 15x2 − 2x − 8
y − 62 = 61x − 61 +62
6x − 20x = −18x
2
4 x
2x − 4
−4x 2 − 24x + 8
−2
2x3 +12x 2 − 4x
−4
2x 3 + 8x 2 − 28x + 8
b. The equation is of the form y = mx + b, and the graph is
a line. So, the equation represents a linear function. c. There are no breaks in the graph. So, it is continuous.
452
Algebra 1 Worked-Out Solutions
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Chapter 7 9. a.
Area of golf hole
Area of Area of = square tee area + rectangular area = x 2 + 3x(x + 4) = x 2 + 3x(x) + 3x(4) = x 2 + 3x 2 + 12x = 4x 2 + 12x
A polynomial that represents the area of the golf hole is ( 4x 2 + 12x ) square feet. b. Perimeter
= x + x + (3x − x) + (x + 4) + 3x + [ (x + 4) + x ] = x + x + 2x + x + 4 + 3x + 2x + 4 = x + x + 2x + x + 3x + 2x + 4 + 4 = 10x + 8 A polynomial that represents the perimeter of the golf hole is (10x + 8) feet. 4x 2 + 12x = 216
c.
4x 2
+ 12x − 216 = 216 − 216
4x 2 + 12x − 216 = 0 4( x 2 + 3x − 54 ) = 0 1 1 —[ 4( x 2 + 3x − 54 ) ] = —(0) 4 4 x 2 + 3x − 54 = 0 Factors of −54
1, 2, −6, −1, −2, −3, 3, 9 54 −54 27 −27 18 −18
6, −9
Sum of factors
53
−25
−3
−53
25
15
−15
3
(x − 6)(x + 9) = 0 x−6=0
or
+6 +6 x=6
x+9=0 −9
−9
x = −9
Disregard x = −9 because a negative side length does not make sense. So, the solution is x = 6. Perimeter = 10x + 8 = 10(6) + 8 = 60 + 8 = 68 So, the perimeter of the golf hole is 68 feet.
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Algebra 1 Worked-Out Solutions
453
