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Solutions Key
CHAPTER
8
Factoring Polynomials
ARE YOU READY? PAGE 521
1. B; a polynomial with two terms
2. A; a whole number greater than 1 that has more
than two whole-number factors
3. F; a number that is multiplied by another number to
get a product
8-1 FACTORS AND GREATEST COMMON
FACTORS, PAGES 524-529
CHECK IT OUT! PAGES 524-526
1a. 40 = 2 · 2 · 2 · 5
= 23 · 5
b. 33 = 3 · 11
2
4. C; the product of any number and a whole number
c. 49 = 7
5. E; a whole number greater than 1 that has exactly
two factors, itself and 1
d. 19 is a prime number.
6. 3, 6, 9, 12
7. 4, 8, 12, 16
8. 8, 16, 24, 32
9. 15, 30, 45, 60
2a. Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 16: 1, 2, 4, 8, 16
The GCF of 12 and 16 is 4.
b. 15 = 3 · 5
25 = 5 · 5
The GCF of 15 and 25 is 5.
10. Yes; 5 × 4 = 20
11. no; factors of 50: 1, 2, 5, 10, 25, 50
12. Yes; 8 × 15 = 120
13. Yes; 7 × 35 = 245
14. Prime
15. Prime
16. Composite;
10 = 2 · 5
17. Composite;
38 = 2 · 19
18. Composite;
115 = 5 · 23
19. Composite;
147 = 21 · 7
20. Prime
21. Prime
22. 2(x + 5)
2(x) + 2(5)
2x +10
23. 3h(h + 1)
3h(h) + 3h(1)
2
3h + 3h
3a. 18g 2 = 2 · 3 · 3 · g · g
27g 3 = 3 · 3 · 3 · g · g · g
The GCF of 18g 2 and 27g 3 is 9g 2.
2
b. 16a = 2 · 2 · 2 · 2 · a · a
9b = 3 · 3 · b
The GCF of 16a 2 and 9b is 1.
c. 8x = 8 · x
2
7v = 7 · v · v
The GCF of 8x and 7v 2 is 1.
4. First find the GCF of 36 and 48 because the number
of CDs per shelves must be a common factor of 36
and 48.
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
The GCF of 36 and 48 is 12.
The greatest possible number of CDs in each shelf
is 12. Find the number of shelves if each shelf holds
12 CDs.
36 CDs by pop artists + 48 CDs by country artists
________________________________________
12 CDs per shelf
= 7 shelves
24. xy(x 2 - xy 3)
xy(x
) - xy(xy 3)
2
3
2 4
x y-x y
25. 6m(m 2 - 4m - 1)
2
6m(m ) - 6m(4m) - 6m(1)
3
6m - 24m 2 - 6m
26. (x + 3)(x + 8)
x(x) + x(8) + 3(x) + 3(8)
x 2 + 8x + 3x + 24
x 2 + 11x + 24
THINK AND DISCUSS, PAGE 526
1. Use a factor tree or divide the number by prime
factors until the quotient is 1.
27. (b - 7)(b + 1)
b(b) + b(1) - 7(b) - 7(1)
b 2 + b - 7b - 7
b 2 - 6b - 7
28. (2p - 5)(p - 1)
2p(p) + 2p(-1) - 5(p) - 5(-1)
2p 2 - 2p - 5p + 5
2p 2 - 7p + 5
2.
x #OEFFICIENT
0RIMEFACTORIZATION
OFCOEFFICIENT
ȕȕȕ
6ARIABLE4ERM
x
6ARIABLETERM
ASAPRODUCT
xȕx
ÊÊ
0RIME
&ACTORIZATION
OFx ȕȕx 29. (3n + 4)(2n + 3)
3n(2n) + 3n(3) + 4(2n) + 4(3)
6n 2 + 9n + 8n + 12
6n 2 + 17n + 12
Copyright © by Holt, Rinehart and Winston.
All rights reserved.
295
Holt Algebra 1
EXERCISES, PAGES 527-529
26. Factors of 14: 1, 2, 7, 14
Factors of 15: 1, 3, 5, 15
The GCF of 14 and 15 is 1.
GUIDED PRACTICE, PAGE 527
1. Possible answer: the greatest number that is a
factor of two given numbers
27. Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30
Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40
The GCF of 30 and 40 is 10.
2
2. 20 = 2 · 2 · 5 = 2 · 5
2
2
3. 36 = 3 · 3 · 2 · 2 = 3 · 2
4. 27 = 3 · 3 · 3 = 3
28. 8a 2 = 2 · 2 · 2 · a · a
11 = 1 · 11
The GCF of 8a 2 and 11 is 1.
3
3
5. 54 = 3 · 3 · 3 · 2 = 3 · 2
29. 9s = 3 · 3 · s
3
63s = 3 · 3 · 7 · s · s · s
The GCF of 9s and 63s 3 is 9s.
5
6. 96 = 2 · 2 · 2 · 2 · 2 · 3 = 2 · 3
7. 7
2
2
8. 100 = 2 · 2 · 5 · 5 = 2 · 5
9. 75 = 3 · 5 · 5 = 3 · 5
4
30. -64n = -1 · 2 · 2 · 2 · 2 · 2 · 2 · n · n · n · n
24n 2 = 2 · 2 · 2 · 3 · n · n
The GCF of -64n 4 and 24n 2 is 8n 2.
2
10. Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 20, 30, 60
The GCF of 12 and 60 is 12.
31. First find the GCF of 72 and 108 because the
number of tarts must be a common factor of 72 and
108.
72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 36, 72
108: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108
To get the greatest possible number of fruits in each
tart, we want the number of tarts to be the least.
The least common factor of 72 and 108 is 2, so
he can make 2 tarts to have 36 raspberries and
72 blueberries in each tart.
11. Factors of 14: 1, 2, 7, 14
Factors of 49: 1, 7, 49
The GCF of 14 and 49 is 7.
12. Factors of 55: 1, 5, 11, 55
Factors of 121: 1, 11, 121
The GCF of 55 and 121 is 11.
13. 6x 2 = 2 · 3 · x · x
5x 2 = 5 · x · x
The GCF of 6x 2 and 5x 2 is x 2.
3
14. 15y = 3 · 5 · y · y · y
-20y = -1 · 2 · 2 · 5 · y
The GCF of 15y 3 and -20y is 5y.
18. 64 = 2 · 2 · 2 · 2 · 2 · 2
= 26
19. 12 = 2 · 2 · 3
= 22 · 3
20. 150 = 2 · 3 · 5 · 5
= 2 · 3 · 52
21. 17
22. 226 = 2 · 113
23. 49 = 7 · 7
24. 63 = 3 · 3 · 7
2
= 32 · 7
=7
25. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 63: 1, 3, 7, 9, 21, 63
The GCF of 36 and 63 is 9.
Copyright © by Holt, Rinehart and Winston.
All rights reserved.
34. 11
35. 2 · n = 2n
37. No; An odd composite number and an even
composite number can have no factors in
common other than 1.
16. First find the GCF of 54 and 18 because the number
of beads per necklace must be a common factor of
54 and 18.
Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54
Factors of 18: 1, 2, 3, 6, 9, 18
The GCF of 54 and 18 is 18.
The greatest possible number of beads in each
necklace is 18. Find the number of necklaces if each
necklace takes 18 beads to make.
54 glass beads + 18 clay beads
__________________________
= 4 necklaces
18 beads per necklace
17. 18 = 2 · 3 · 3
= 2 · 32
33. 2 · 2 · x · x = 4x 2
36. Possible answer: Even numbers greater than 2 all
have 2 as a factor and thus are not prime.
4
15. 13q = 13 · q · q · q · q
2
2p = 2 · p · p
The GCF of 13q 4 and 2p 2 is 1.
PRACTICE AND PROBLEM SOLVING, PAGES 527–528
32. 5 · t = 5t
38a. Since the area of a rectangle is length times width,
to find all possible whole number lengths, find
2 whole numbers that have the product 84.
1 × 84; 2 × 42; 3 × 28; 4 × 21; 6 × 14; 7 × 12
b. P = 2(7 + 12)
= 2(19) = 38 ft
c. P = 2(1 + 84)
= 2(85) = 170 ft
39. First find the GCF of 35 and 40, because the
number of guards in each row must be a common
factor of 35 and 40.
35: 1, 5, 7, 35
40: 1, 2, 4, 5, 8, 10, 20, 40
The GCF of 35 and 40 is 5.
The greatest possible number of guards in each
row is 5. Find the number of rows if each row has
5 guards.
35 Cavaliers + 40 Blue Devils
_________________________
= 15 rows
5 Guards per row
41. 8: 1, 2, 4, 8
40. 11: 1, 11
20: 1, 2, 4, 5, 10, 20
12: 1, 2, 3, 4, 6, 12
63: 1, 3, 7, 9, 21, 63
14: 1, 2, 7, 14
8 and 20; GCF = 4
12 and 14; GCF = 2
296
Holt Algebra 1
42. 16: 1, 2, 4, 8, 16
21: 1, 3, 7, 21
27: 1, 3, 9, 27
21 and 27; GCF = 3
43. 32: 1, 2, 4, 8, 16, 32
63: 1, 3, 7, 9, 21, 63
105: 1, 3, 5, 7, 15, 21,
35, 105
63 and 105; GCF = 21
44. 25: 1, 5, 25
35: 1, 5, 7, 35
54: 1, 2, 3, 4, 6, 9, 16,18, 27, 54
25 and 35; GCF = 5
45. 35: 1, 5, 7, 35
54: 1, 2, 3, 6, 9, 18, 27, 54
72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
54 and 72; GCF = 18
4
46. 2 · 3; possible answer: because 48 = 2 · 24 and
24 = 2 3 · 3, 48 = 2 · 2 3 · 3 = 2 4 · 3
From top to bottom, left to right:
47. 36; 2; 9; 3
48. 27; 3; 9
49. 105; 5; 7
50. 2; 14; 7
51. 2; 2; 27; 3
52. 2; 34; 17
53. 24; 2; 6; 3
54. 2; 70; 5
55. 2; 2; 10; 5; 2 3 × 18
1 at 2
56a. Use the given formula, d = vt + __
2
1 (2)t 2
d = (2)t + __
2
2
= 2t + t
b. 2t = 2 · t
2
t =t·t
The GCF of 2t and t 2 is t.
62. 100 = 2 · 2 · 5 · 5
25s 5 = 5 · 5 · s · s · s · s · s
50s = 2 · 5 · 5 · s
The GCF of 100, 25s 5, and 50s is 25.
4
63. 2p r = 2 · p · p · p · p · r
8p 3r 2 = 2 · 2 · 2 · p · p · p · r · r
16p 2r 3 = 2 · 2 · 2 · 2 · p · p · r · r · r
The GCF of 2p 4r, 8p 3r 2, and 16p 2r 3 is 2p 2r.
3
64. 2x y = 2 · x · x · x · y
8x 2y 2 = 2 · 2 · 2 · x · x · y · y
17xy 3 = 17 · x · y · y · y
The GCF of 2x 3y, 8x 2y 2, and 17xy 3 is xy.
4 3
65. 8a b = 2 · 2 · 2 · a · a · a · a · b · b · b
4a 3b 3 = 2 · 2 · a · a · a · b · b · b
12a 2b 3 = 2 · 2 · 3 · a · a · b · b · b
The GCF of 8a 4b 3, 4a 3b 3, and 12a 2b 3 is 4a 2b 3.
66. 1 × 20; 2 × 10; 4 × 5; 20 × 1; 10 × 2; 5 × 4
67. Possible answer: 21, 35, 49; 7, 14, 84
68. Possible answer: 6, 35, 143
6 = 2 × 3; 35 = 5 × 7; 143 = 11 × 13
SPIRAL REVIEW, PAGE 529
69. 40% of 60
70. 250% of 16
= 0.4 × 60 = 24
= 2.5 × 16 = 40
20
___
71.
= 0.25 = 25%
80
72. Yes; common difference = 4; 19, 23, 27
73. No
TEST PREP, PAGE 529
74. Yes; common difference = -0.5; -0.5, -1.0, -1.5
57. D; 16, 24, 48 has a GCF of 8.
2
2
75. P = (x + 3x) + (6x - 2) + (2x + 5x - 1)
2
= 3x + 14x - 3
58. F; the GCF of 48 and 12 is 12, and the GCF of 12
and 8 is 4.
FT
59.
FT
P = 50 ft
MODEL FACTORING, PAGE 530
TRY THIS, PAGE 530
FT
1. 3(x + 3)
FT
P = 28 ft
FT
FT
P = 22 ft
•
2. 2(x + 4)
FT
FT
P = 20 ft
Patricia should make the pen 4 ft × 6 ft because
these dimensions give the shortest perimeter and
she will need to buy the least fencing.
3. 4(x - 3)
CHALLENGE AND EXTEND, PAGE 529
3
60. 4n = 2 · 2 · n · n · n
16n 2 = 2 · 2 · 2 · 2 · n · n
8n = 2 · 2 · 2 · n
The GCF of 4n 3, 16n 2, and 8n is 4n.
•
4. 3(x - 4)
3
61. 27y = 3 · 3 · 3 · y · y · y
18y 2 = 2 · 3 · 3 · y · y
81y = 3 · 3 · 3 · 3 · y
The GCF of 27y 3, 18y 2, and 81y is 9y.
Copyright © by Holt, Rinehart and Winston.
All rights reserved.
•
297
•
Holt Algebra 1
c. 3x(y + 4) - 2y(x + 4)
There are no common factors.
5. 2x(x + 1)
•
d. 5x(5x - 2) - 2(5x - 2)
(5x - 2)(5x - 2)
(5x - 2)2
3
2
4a. 6b + 8b + 9b + 12
(6b 3 + 8b 2) + (9b + 12)
2b 2(3b + 4) + 3(3b + 4)
2
(2b + 3)(3b + 4)
6. x(x + 4)
•
3
2
b. 4r + 24r + r + 6
3
(4r + 24r) + (r 2 + 6)
4r(r 2 + 6) + 1(r 2 + 6)
(4r + 1)(r 2 + 6)
7. x(x - 3)
•
2
3
5a. 15x - 10x + 8x - 12
3
-10x + 15x 2 + 8x - 12
(-10x 3 + 15x 2) + (8x - 12)
-5x 2(2x - 3) + 4(2x - 3)
(-5x 2 + 4)(2x - 3)
8. 2x(x - 2)
•
b. 8y - 8 - xy + x
8y - xy - 8 + x
y(8 - x) - 1(8 - x)
(y - 1)(8 - x)
8-2 FACTORING BY GCF, PAGES 531-538
CHECK IT OUT! PAGES 532–534
THINK AND DISCUSS, PAGE 534
1. Possible answer: when you know the GCF of the
monomials in a polynomial, you can factor out the
GCF from each monomial to factor the polynomial.
1a. 5b = 2 · b
9b 3 = 3 · 3 · b · b · b
The GCF of 5b and 9b 3 is b.
5(b) + 9b 2(b)
b(5 + 9b 2)
2.
&ACTORINGBY'#&
2
b. 9d = 3 · 3 · d · d
82 = 2 · 2 · 2 · 2 · 2 · 2
The GCF of 9d 2 and 8 2 is 1; it cannot be factored.
&INDTHEGREATESTCOMMONFACTOR
7RITEEACHTERMASAPRODUCTUSING
THE'#&
3
c. 18y = 2 · 3 · 3 · y · y · y
2
7y = 7 · y · y
The GCF of 18y 3 and 7y 2 is y 2.
-1[18y(y 2) + 7(y 2)]
-y 2(18y + 7)
4
d. 8x = 2 · 2 · 2 · x · x · x · x
4x 3 = 2 · 2 · x · x · x
2x 2 = 2 · x · x
The GCF of 8x 4, 4x 3 and 2x 2 is 2x 2
4x 2(2x 2) + 2x(2x 2) - 1(2x 2)
2x 2(4x 2 + 2x - 1)
5SETHE$ISTRIBUTIVE0ROPERTYTOFACTOR
OUTTHE'#&
#HECKBYMULTIPLYING
EXERCISES, PAGES 535–537
GUIDED PRACTICE, PAGE 535
1. 15a = 3 · 5 · a
5a 2 = 5 · a · a
The GCF of 15a and 5a 2 is 5a.
15a - 5a 2
3(5a) - a(5a)
5a(3 - a)
2
2. A = 2x + 4x
= x(2x) + 2(2x)
= 2x(x + 2)
Possible expression for the dimensions of the solar
panel are 2x cm and (x + 2) cm.
2. 10g 3 = 2 · 5 · g · g · g
3g = 3 · g
The GCF of 10g 3 and 3g is g.
10g 3 - 3g
10g 2(g) - 3(g)
g(10g 2 - 3)
3a. 4s(s + 6) - 5(s + 6)
(4s - 5)(s + 6)
b. 7x(2x + 3) + (2x + 3)
7x(2x + 3) + 1(2x + 3)
(7x + 1)(2x + 3)
Copyright © by Holt, Rinehart and Winston.
All rights reserved.
298
Holt Algebra 1