1. Adding and Subtracting Polynomials
ALGEBRA 1 LESSON 9-1
(For help, go to Lesson 1-7.)
Simplify each expression.
1. 6t + 13t 2. 5g + 34g
3. 7k – 15k 4. 2b – 6 + 9b
5. 4n2 – 7n2 6. 8x2 – x2
9-1

2. Adding and Subtracting Polynomials
ALGEBRA 1 LESSON 9-1
Solutions
1. 6t + 13t = (6 + 13)t = 19t 2. 5g + 34g = (5 + 34)g = 39g
3. 7k – 15k = (7 – 15)k = –8k 4. 2b – 6 + 9b = (2 + 9)b – 6 = 11b – 6
5. 4n2 – 7n2 = (4 – 7)n2 = –3n2 6. 8x2 – x2 = (8 – 1)x2 = 7x2
9-1

3. Adding and Subtracting Polynomials
ALGEBRA 1 LESSON 9-1
Find the degree of each monomial.
a. 18
The degree of a nonzero constant is 0.
Degree: 0
b. 3xy3
The exponents are 1 and 3. Their sum is 4.
Degree: 4
c. 6c
6c = 6c1. The exponent is 1.
Degree: 1
9-1

4. Adding and Subtracting Polynomials
ALGEBRA 1 LESSON 9-1
Write each polynomial in standard form. Then name each polynomial by its degree and the number of its terms.
a. –2 + 7x
Place terms in order.
7x – 2
linear binomial
b. 3x5 – 2 – 2x5 + 7x
Place terms in order.
3x5 – 2x5 + 7x – 2
x5 + 7x – 2
Combine like terms.
fifth degree trinomial
9-1

5. Adding and Subtracting Polynomials
ALGEBRA 1 LESSON 9-1
Simplify (6x2 + 3x + 7) + (2x2 – 6x – 4).
Method 1: Add vertically.
Line up like terms. Then add the coefficients.
6x2 + 3x + 7
2x2 – 6x – 4
8x2 – 3x + 3
Method 2: Add horizontally.
Group like terms. Then add the coefficients.
(6x2 + 3x + 7) + (2x2 – 6x – 4) = (6x2 + 2x2) + (3x – 6x) + (7 – 4)
= 8x2 – 3x + 3
9-1

6. Adding and Subtracting Polynomials
ALGEBRA 1 LESSON 9-1
Simplify (2x3 + 4x2 – 6) – (5x3 + 2x – 2).
Method 1: Subtract vertically.
Line up like terms. Then add the coefficients.
(2x3 + 4x2 – 6) Line up like terms.
–(5x3 – 2x – 2)
2x3 + 4x2 – 6 Add the opposite.
–5x – 2x + 2
–3x3 + 4x2 – 2x – 4
9-1

7. Adding and Subtracting Polynomials
ALGEBRA 1 LESSON 9-1
(continued)
Method 2: Subtract horizontally.
(2x3 + 4x2 – 6) – (5x3 + 2x – 2)
= 2x3 + 4x2 – 6 – 5x3 – 2x + 2 Write the opposite of each term in the polynomial being subtracted.
= (2x3 – 5x3) + 4x2 – 2x + (–6 + 2) Group like terms.
= –3x3 + 4x2 – 2x – 4 Simplify.
9-1

8. Adding and Subtracting Polynomials
ALGEBRA 1 LESSON 9-1
pages 459–461  Exercises
1. 1
2. 3
3. 0
4. 10
5. 4
6. 0
7. 4
8. 0
9. quadratic trinomial
10. linear binomial
11. cubic trinomial
12. not a polynomial
13. constant monomial
14. quadratic binomial
15. –3x2 + 4x;
quadratic binomial
16. 4x + 9; linear binomial
17. c2 + 4c – 2;
quadratic trinomial
18. –2z2 + 5z – 5;
19. 15y8 – 7y3 + y;
eighth degree trinomial
20. 4q4 + 3q2 – 8q – 10;
fourth degree polynomial with 4 terms
21. 8m2 + 15
22. 10k + 4
23. 8w2 – 3w + 4
24. 20x2 + 7
25. 10g4 + 11g
26. 6a2 – 7a + 21
27. 8y4 + 7y3 + 4y
28. 2c – 14
29. b + 1
30. 4h2 + 6h – 18
9-1

9. Adding and Subtracting Polynomials
ALGEBRA 1 LESSON 9-1
31. 7n4 + n3
32. 15x5 + x
33. 5w2 – 4w + 10
34. –5x4 – x3 – 7x2 + x
35. 18y2 + 5y
36. –6x3 + 3x2 – 4
37. –7z3 + 6z2 + 2z – 5
38. 7a3 + 11a2 – 4a – 4
39. 28c – 16
40. 39x – 7
41. Kwan did not take the
opposite of each term in
the polynomial being subtracted.
42. a. monogram: a design composed of one
or more letters, typically the initials of a
name; used as an identifying mark
binocular: relating to, used by, or
involving both eyes at the same time
tricuspid: having three cusps, usually said of a molar tooth
polyglot: a person with a speaking, reading, or writing knowledge of several languages
b. Answers may vary. Sample: monopoly,
biathlon, tripod, polychrome
c. yes
43. –x4 + x3 + 15x
44. –7g8 – 2g3 – 11g2 + 2g
9-1

10. Adding and Subtracting Polynomials
ALGEBRA 1 LESSON 9-1
45. –h10 – 5h9 + 8h5 + 2h4
46. t4 + 4t + 6
47. –5b9 + 3b8 – 4b7 + 8b
48. –5k4 + k3 – k2 + 11
49. 5x + 18
50. 9a – 6
51. No; both terms of a binomial cannot be constants.
52. a. y = 2x – 1; y = 0.5x + 3
b. D(x) = 1.5x – 4
c. or 2
d. The lines intersect at x = 2 .
53. 5a3b – 5ab
54. –p4q4 – 11p4q + 14pq6
55. a. p(t) = 57t
b. 2,617,300
c. the difference between the number of
men and the number of women enrolled
in a college
56. D
57. G
58. A
59. C
60. B
61. C
62. [2] (9x3 – 4x2 + 1) – (x2 + 2) =
9x3 – 4x2 + 1 – x2 – 2 =
(9x3) + (–4x2 – x2 ) + (1 – 2) =
9x3 – 5x2 – 1
[1] one incorrect term OR no work shown 8 3 2 3 2 3
9-1

11. Adding and Subtracting Polynomials
ALGEBRA 1 LESSON 9-1
63. 2
65. 5
66. 5
67. exponential growth
68. exponential decay
69. exponential growth
70. exponential decay
72. 2–1
73. 36x5
74. 3a3b2
75. –30t–7
76. (–3)2
77. –12h10
78. 10q7
79. y = |x| + 5
80. y = |x – 6|
81. y = |x| – 12
82. y = |x| + 7
83. y = |x + 10|
84. y = |x + 0.4|
85. y = |x| + 5.2
86. y = |x| – 2.3
9-1

12. Adding and Subtracting Polynomials
ALGEBRA 1 LESSON 9-1
Simplify each expression. Then name each polynomial by its degree and number of terms.
1. –4 + 3x – 2x2
2. 2b2 – 4b3 + 6
3. (2x4 + 3x – 4) + (–3x x4)
4. (–3r + 4r2 – 3) – (4r2 + 6r – 2)
–2x2 + 3x – 4; quadratic trinomial
–4b3 + 2b2 + 6; cubic trinomial
3x4; fourth degree monomial
–9r – 1; linear binomial
9-1

13. Multiplying and Factoring
ALGEBRA 1 LESSON 9-2
(For help, go to Lesson 1–7.)
Multiply.
1. 3(302) 2. 41(7) 3. 9(504)
Simplify each expression.
4. 4(6 + 5x) 5. –8(2y + 1)
6. (5v – 1)5 7. 7(p – 2)
8. (6 – x)9 9. –2(4q – 1)
9-2

14. Multiplying and Factoring
ALGEBRA 1 LESSON 9-2
Solutions
1. 3(302) = (7) = (504) = 4536
4. 4(6 + 5x) = 4(6) + 4(5x) = x
5. –8(2y + 1) = (–8)(2y) + (–8)(1) = –16y – 8
6. (5v – 1)5 = (5v)(5) – (1)(5) = 25v – 5
7. 7(p – 2) = 7p – 7(2) = 7p – 14
8. (6 – x)9 = 6(9) – 9x = 54 – 9x
9. –2(4q – 1) = (–2)(4q) – (–2)(1) = –8q + 2
9-2

15. Multiplying and Factoring
ALGEBRA 1 LESSON 9-2
Simplify –2g2(3g3 + 6g – 5).
–2g2(3g3 + 6g – 5)
= –2g2(3g3) –2g2(6g) –2g2(–5) Use the Distributive Property.
= –6g2 + 3 – 12g g2 Multiply the coefficients and add the exponents of powers with the same base.
= –6g5 – 12g3 + 10g2 Simplify.
9-2

16. Multiplying and Factoring
ALGEBRA 1 LESSON 9-2
Find the GCF of 2x4 + 10x2 – 6x.
List the prime factors of each term. Identify the factors common to all terms.
2x4 = 2 • x • x • x • x
10x2 = 2 • 5 • x • x
6x = 2 • 3 • x
The GCF is 2 • x, or 2x.
9-2

17. Multiplying and Factoring
ALGEBRA 1 LESSON 9-2
Factor 4x3 + 12x2 – 16x.
Step 1: Find the GCF.
Step 2: Factor out the GCF.
4x3 + 12x2 – 16x
4x3 = 2 • 2 • x • x • x
12x2 = 2 • 2 • 3 • x • x
16x = 2 • 2 • 2 • 2 • x
= 4x(x2) + 4x(3x) + 4x(–4)
= 4x(x2 + 3x – 4)
The GCF is 2 • 2 • x, or 4x.
9-2

18. Multiplying and Factoring
ALGEBRA 1 LESSON 9-2
pages 463–465  Exercises
1. 8m2 + 48m
2. 3x2 + 30x
3. 63k2 + 36k
4. –5a2 + 5a
5. 18x2 + 2x3
6. –p3 + 11p2
7. 12x4 – 2x3 + 10x2
8. 36y5 + 32y4 – 44y2
9. –45c5 + 40c4 + 25c3
10. –42q7 + 14q3 + 49q2
11. –3g g9 – 15g7
12. –40x9 – 12x8 + 28x6
13. 3
14. 2a
15. 12
16. x
17. 5
18. 3x
19. 2(3x – 2)
20. v(v + 4)
21. 5(2x3 – 5x2 + 4)
22. 2t2(1 – 5t2)
23. 3n(5n2 – n + 4)
24. 6p3(p3 + 4p2 + 3)
25. Karla; Kevin multiplied
–2x by 3 instead of –3.
26. Answers may vary.
Sample: 8x3 + 12x2 + 24x;
4x(2x2 + 3x + 6)
27. –12a3 + 15a2 – 27a
28. 14p5 – 35p3
29. –60c3 + 36c2 – 48c
30. –4y2 + 13y
31. x2 + x
32. 12t 3 – 23t 2
33. a. A = 16 x2 – 4x2
b. A = 4x2(4 – 1)
9-2

19. Multiplying and Factoring
ALGEBRA 1 LESSON 9-2
34. 9m5(m7 – 4m2 + 9)
35. 24x(x2 – 4x + 2)
36. 16n(n2 + 3n – 5)
37. x2(5x2 + 4x + 3)
38. 13ab3(1 + 3ab)
39. 7g 2k2(k – 5g3)
40. 25; 52 = 25
41. a. n(n – 1)
b. Always; the product of
two consecutive integers
is always even since one
of the integers is even.
42. a. 1; 2; 3; 4; (n + 1)
b. 5050
43. a. 6; 3
b. n – 3
c. n2 – n
d. 20
44. a. V = 64s3
b. V = s2
c. V = 64s3 – s2
d. V = 16s2(4s – 3 )
e. about 182,071 in.3
45. B
46. F
47. C
48. G
49. B 1 2 3 2 n 2
9-2

20. Multiplying and Factoring
ALGEBRA 1 LESSON 9-2
57.
58.
59. –
60. 1
61.
62.
63. 4c3
64.
65. (3, 2)
66. (5, –3)
67. (1, 6)
50. [2] 2x2(5x2 + 3x + 1);
the terms 5x2, 3x,
and 1 have no
common factor
other than 1.
[1] incorrect factoring
or incorrect
explanation
51. –3x2 + 10
52. m3 – 5m2 + 12m + 6
53. 5g 2 – g
54. r2 – 16r + 16
55. t4 + 5t 2 – 9
56. 3b3 – 6b2 – 12 1 5 1 25 1 8 m2 n3
v 3 w5 a
b8c5
9-2

21. Multiplying and Factoring
ALGEBRA 1 LESSON 9-2
1. Simplify –2x2(–3x2 + 2x + 8).
2. Find the GCF of 16b4 – 4b3 + 8b2.
3. Factor 3x3 + 9x2.
4. Factor 10y3 + 5y2 – 15y.
6x4 – 4x3 – 16x2
4b2
3x2(x + 3)
5y(2y + 3)(y – 1)
9-2

22. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
(For help, go to Lesson 9-2.)
Find each product.
1. 4r(r – 1) 2. 6h(h2 + 8h – 3) 3. y2(2y3 – 7)
Simplify. Write each answer in standard form.
4. (x3 + 3x2 + x) + (5x2 + x + 1)
5. (3t3 – 6t + 8) + (5t3 + 7t – 2)
6. w(w + 1) + 4w(w – 7) 7. 6b(b – 2) – b(8b + 3)
8. m(4m2 – 6) + 3m2(m + 9) 9. 3d2(d3 – 6) – d3(2d2 + 4)
9-3

23. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
Solutions
1. 4r(r – 1) = 4r(r) – 4r(1) = 4r 2 – 4r
2. 6h(h2 + 8h – 3) = 6h(h2) + 6h(8h) – 6h(3) = 6h3 + 48h2 – 18h
3. y2(2y3 – 7) = y2(2y3) – 7y2 = 2y5 – 7y2
4. x3 + 3x2 + x 5. 3t3 – 6t + 8
x2 + x t3 + 7t – 2
x3 + 8x2 + 2x t3 + t + 6
6. w(w + 1) + 4w(w – 7) 7. 6b(b – 2) – b(8b + 3)
= w(w) + w(1) + 4w(w) – 4w(7) = 6b(b) – 6b(2) – b(8b) – b(3)
= w2 + w + 4w2 – 28w = 6b2 – 12b – 8b2 – 3b
= (1 + 4)w2 + (1 – 28)w = (6 – 8)b2 + (–12 – 3)b
= 5w2 – 27w = –2b2 – 15b
9-3

24. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
Solutions (continued)
8. m(4m2 – 6) + 3m2(m + 9)
= m(4m2) – m(6) + 3m2(m) + 3m2(9)
= 4m3 – 6m + 3m3 + 27m2
= (4 + 3)m3 + 27m2 – 6m
= 7m3 + 27m2 – 6m
9. 3d2(d3 – 6) – d3(2d2 + 4)
= 3d2(d3) – 3d2(6) – d3(2d2) – d3(4)
= 3d5 – 18d2 – 2d5 – 4d3
= (3 – 2)d5 – 4d3 – 18d2
= d5 – 4d3 – 18d2
9-3

25. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
Simplify (2y – 3)(y + 2).
(2y – 3)(y + 2) = (2y – 3)(y) + (2y – 3)(2) Distribute 2y – 3.
= 2y2 – 3y + 4y – 6 Now distribute y and 2.
= 2y2 + y – 6 Simplify.
9-3

26. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
Simplify (4x + 2)(3x – 6).
First
= (4x)(3x)
(4x + 2)(3x – 6)
Outer
(4x)(–6) +
Inner
(2)(3x) +
Last
(2)(–6) +
24x 6x 12 – +
= 12x2
= 12x2
18x – 12
The product is 12x2 – 18x – 12.
9-3

27. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
Find the area of the shaded region. Simplify.
area of outer rectangle = (3x + 2)(2x – 1)
area of hole = x(x + 3)
area of shaded region = area of outer rectangle – area of hole
= (3x + 2)(2x – 1) –x(x + 3) Substitute.
= 6x2 – 3x + 4x – 2 –x2 – 3x Use FOIL to simplify (3x + 2) (2x – 1) and the Distributive Property to simplify x(x + 3).
= 6x2 – x2 – 3x + 4x – 3x – 2 Group like terms.
= 5x2 – 2x – 2 Simplify.
9-3

28. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
Simplify the product (3x2 – 2x + 3)(2x + 7).
Method 1: Multiply using the vertical method.
3x2  –   2x  +  3
2x  +  7
21x2  –  14x  +  21   Multiply by 7.
6x3  –  4x2  +   6x Multiply by 2x.
6x3  + 17x2  –   8x  +  21 Add like terms.
9-3

29. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
(continued)
Method 2: Multiply using the horizontal method.
= (2x)(3x2) – (2x)(2x) + (2x)(3) + (7)(3x2) – (7)(2x) + (7)(3)
(2x + 7)(3x2 – 2x + 3)
= 6x3 – 4x2 + 6x + 21x2 – 14x + 21
= 6x3 + 17x2 – 8x + 21
The product is 6x3 + 17x2 – 8x + 21.
9-3

30. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
pages 469–472  Exercises
1. 30
2. 35
3. 7
4. 13
5. x2 + 7x + 10
6. h2 + 7h + 12
7. k2 + k – 42
8. a2 – 17a + 72
9. 2x2 + 3x – 2
10. 2y2 – y – 15
11. r 2 + 2r – 24
12. 5y2 + 12y – 32
13. x2 – x – 42
14. m2 – 15m + 54
15. 4b2 + 10b – 6
16. 8w2 + 42w + 10
17. x2 + 2x – 63
18. a2 + 16a + 55
19. p2 + 9p – 10
20. 8x + 6
21. –2x2 + 5x + 48
22. x3 + 5x2 – 35x + 9
23. a3 – 6a2 + 9a – 4
24. 2g 3 – 3g 2 – 6g – 9
25. 3k3 + 19k2 – 33k + 56
26. 9x3 + 15x2 + 3x – 3
27. 2t 3 – 17t t – 15
28. 56p p2 + 37p – 9
29. 48w3 – 28w2 – 2w + 2
30. p2 + p – 56
31. p2 + p – 56
32. p3 + 8p2 – 7p – 56
33. 25c2 – 40c – 9
34. n3 + 11n2 + 3n + 33
35. 15k4 + 3k3 + 10k2 + 2k
9-3

31. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
36. 24h3 + 2h2 + 17h – 3
37. 9y4 – 9y3 – 7y2 – 2y – 2
38. 48q3 – 16q2 + 4q – 4
38. a. 2x2 + 12x + 16
b. 12x + 16
c. 10 ft by 5 ft
40. Answers may vary. Sample:
(x + 2)(x2 + 3x + 4);
x3 + 5x2 + 10x + 8
41. Answers may vary. Sample:
vertical method, so you can
keep terms aligned
x + 15
x x – 1
44. a. x2 + 2x + 1, 121
x2 + 3x + 2, 132
x2 + 4x + 3, 143
b. For ax2 + bx + c, a corresponds
to the hundreds place, b to the
tens place, and c to the ones place.
45. n3 + 15n2 + 56n
46. n2 + 6n + 8
47. 6x2 + 36x + 54
48. 96t t + 6
49. 24w w
50. a. V(t) = 7.02t t + 40,920
b. 46,588.5 million lb
51. a. 2000r r r
b. $
9-3

32. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
g g + 4
k k + 25
x x +
59. A
60. H
61. B
62. [2] 8v3 + 2v2 + 3v – 1; multiply each term of
(2v2 + v + 1) by 4v, and then multiply each term of (2v2 + v + 1) by –1. Add the 6 products.
[1] incorrect calculation OR incorrect explanation
63. [4] [(2n + 6)(5n – 2)] – [(3n + 2)(n + 2)] =
[10n2 + 26n – 12] – [3n2 + 8n + 4] =
7n2 + 18n – 16
[3] appropriate methods, but with one
computational error
[2] incorrect products subtracted correctly
OR correct products subtracted incorrectly
[1] correct answer, without work shown
64. 20v2 – 28v
65. 3c2 – 27c
66. 8t t 2
9-3

33. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
67. 3y2 – 10y
68. –5x3 + 55x2
69. –6t4 + t3
70. –4r6 + 12r
71. 9b5 + 18b3
72. 5(w + 9)
73. x(3x – 11)
74. 4a(a + 3)
75. n2(9 – n)
76. 17(2t – 3)
77. 9v(7v + 5)
78. 5m(5 – 12m2)
79. 11k(1 + 7k5)
80. 27
81.
82. y4
83.
84.
85.
86.
87. 1
88.
89. 64y4 1 27 1
3w5 1
2x11 3 5 9 25 49
16x2
9-3

34. Multiplying Binomials
ALGEBRA 1 LESSON 9-3
Simplify each product using any method.
1. (x + 3)(x – 6) 2. (2b – 4)(3b – 5)
3. (3x – 4)(3x2 + x + 2)
4. Find the area of the shaded region.
x2 – 3x – 18
6b2 – 22b + 20
9x3 – 9x2 + 2x – 8
2x2 + 3x – 1
9-3

35. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
(For help, go to Lessons 8–4 and 9-3.)
Simplify.
1. (7x)2 2. (3v)2 3. (–4c)2 4. (5g3)2
Use FOIL to find each product.
5. (j + 5)(j + 7) 6. (2b – 6)(3b – 8)
7. (4y + 1)(5y – 2) 8. (x + 3)(x – 4)
9. (8c2 + 2)(c2 – 10) 10. (6y2 – 3)(9y2 + 1)
9-4

36. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
Solutions
1. (7x)2 = 72 • x2 = 49x2 2. (3v)2 = 32 • v2 = 9v2
3. (–4c)2 = (–4)2 • c2 = 16c2 4. (5g3)2 = 52 • (g3)2 = 25g6
5. (j + 5)(j + 7) = (j)(j) + (j)(7) + (5)(j) + (5)(7)
= j2 + 7j + 5j + 35
= j2 + 12j + 35
6. (2b – 6)(3b – 8) = (2b)(3b) + (2b)(–8) + (–6)(3b) + (–6)(–8)
= 6b2 – 16b – 18b + 48
= 6b2 – 34b + 48
9-4

37. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
Solutions (continued)
7. (4y + 1)(5y – 2)) = (4y)(5y) + (4y)(–2) + (1)(5y) + (1)(–2)
= 20y2 – 8y + 5y – 2
= 20y2 – 3y – 2
8. (x + 3)(x – 4) = (x)(x) + (x)(-4) + (3)(x) + (3)(–4)
= x2 – 4x + 3x – 12
= x2 – x – 12
9. (8c2 + 2)(c2 – 10) = (8c2)(c2) + (8c2)(–10) + (2)(c2) + (2)(–10)
= 8c4 – 80c2 + 2c2 – 20
= 8c4 – 78c2 – 20
10. (6y2 – 3)(9y2 + 1) = (6y2)(9y2) + (6y2)(1) + (–3)(9y2) + (–3)(1)
= 54y4 + 6y2 – 27y2 – 3
= 54y4 – 21y2 – 3
9-4

38. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
a. Find (y + 11)2.
(y + 11)2 = y2 + 2y(11) + 72 Square the binomial.
= y2 + 22y Simplify.
b. Find (3w – 6)2.
(3w – 6)2 = (3w)2 –2(3w)(6) + 62 Square the binomial.
= 9w2 – 36w + 36 Simplify.
9-4

39. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
Among guinea pigs, the black fur gene (B) is dominant and the white fur gene (W) is recessive. This means that a guinea pig with at least one dominant gene (BB or BW) will have black fur. A guinea pig with two recessive genes (WW) will have white fur.
The Punnett square below models the possible combinations of color genes that parents who carry both genes can pass on to their offspring. Since WW is of the outcomes, the probability that a guinea pig has white fur is . 1 4
BB BW
BW WW B W
B W
9-4

40. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
(continued)
You can model the probabilities found in the Punnett square with the expression ( B + W)2. Show that this product gives the same result as the Punnett square. 1 2
( B + W)2 = ( B)2 – 2( B)( W) + ( W)2 Square the binomial. 1 2
= B BW + W 2 Simplify. 1 4 2
The expressions B2 and W 2 indicate the probability that offspring will have either two dominant genes or two recessive genes is . The expression BW indicates that there is chance that the offspring will inherit both genes. These are the same probabilities shown in the Punnett square. 1 4 2
9-4

41. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
a. Find 812 using mental math.
812 = (80 + 1)2
= (80 • 1) + 12 Square the binomial.
= = 6561 Simplify.
b. Find 592 using mental math.
592 = (60 – 1)2
= 602 – 2(60 • 1) + 12 Square the binomial.
= 3600 – = 3481 Simplify.
9-4

42. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
Find (p4 – 8)(p4 + 8).
(p4 – 8)(p4 + 8) = (p4)2 – (8)2 Find the difference of squares.
= p8 – 64 Simplify.
9-4

43. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
Find 43 • 37.
43 • 37 = (40 + 3)(40 – 3) Express each factor using 40 and 3.
= 402 – 32 Find the difference of squares.
= 1600 – 9 = 1591 Simplify.
9-4

44. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
pages 477–479  Exercises
1. c2 + 2c + 1
2. x2 + 8x + 16
3. 4v2 + 44v + 121
4. 9m2 + 42m + 49
5. w 2 – 24w + 144
6. b2 – 10b + 25
7. 36x2 – 96x + 64
8. 81j 2 – 36j + 4
9. a. C CD D2 b.
c. It is the coefficient of C2.
13. 91,204
,001
15. x2 – 16
16. a2 – 64
17. d 2 – 49
18. h2 – 225
19. y2 – 144
20. k2 – 25
24. 39,991
25. 89,999
26. (6x + 9) units2
27. (10x + 15) units2
28. x2 + 6xy + 9y2
29. 25p2 – 10pq + q2
30. 36m2 + 12mn + n2
31. x2 – 14xy + 49y2
32. 16k2 + 56kj + 49j 2
33. 4y2 – 36xy + 81x2 1 16 3 8 9 16 1 16
9-4

45. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
34. 9w wt + 100t 2
35. 36a ab + 121b2
36. 25p2 – 60pq + 36q 2
37. 36h2 – 96hp + 64p2
38. y10 – 18x4y5 + 81x8
39. 64k kh + 16h2
40. a. R + W = R RW + W 2 b.
c. R + W R = R RW
d. 0
41. a.
b. n2 is one more than the
product (n – 1)(n + 1).
c. The product (n – 1)(n + 1)
is n2 – 1.
42. Answers may vary. Sample:
(2 + 2)
16 = 8
43. No; = = =
32 + 2(3) = = = 9 1 2 1 2 1 2 2 1 4 1 2 1 4 1 2 / 1 2 1 2 1 2 1 2 1 2 1 2 1 2 2 1 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 2 1 4 1 2 1 4 /
9-4

46. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
44. 9y2 – 25w2
45. p2 – 81q2
46. 4d2 – 49g2
47. 49b2 – 64c2
48. g2 – 49h2
49. g6 – 49h4
50. 4a4 – b2
x2 – y6
52. 16k2 – 9h4
53. a2 + b2 + c2 + 2ab + 2bc + 2ac
54. a. H H 2T + HT T 3 b.
55. a. (3n + 1)2 = (3n + 1)(3n + 1) =
9n2 + 6n + 1 = 3(3n2 + 2n) + 1;
since 3n2 + 2n is an integer, then
3(3n2 + 2n) is a multiple of three
and 3(3n2 + 2n) + 1 is one more
than a multiple of three.
b. No; its square is one more than
a multiple of three.
56. V = r r r + 36
57. a. b. 4 3 1 8 3 8 3 8 1 8 3 8 3 8
9-4

47. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
58. D
59. F
60. C
61. B
62. C
63. [2] The middle term is twice
the product of the first and
last terms; 2(3x)(–4y) = –24xy.
[1] incorrect explanation
64. k2 – 2k – 63
65. 2x2 – 23x + 66
66. 15p2 + 7p – 4
67. 3y2 + 4y + 1
68. 24h2 – 8h – 2
69. 72b2 + 74b + 14
70. 2w w 2 + 5w + 40
71. r 3 – 4r 2 – 30r + 63
72. 30m5 + 8m3 – 8m
 103
 10–2
 104
 106
 101
 102
79. 6  109
 10–1
9-4

48. Multiplying Special Cases
ALGEBRA 1 LESSON 9-4
Find each square.
1. (y + 9)2 2. (2h – 7)2
5. Find (p3 – 7)(p3 + 7). 6. Find 32 • 28.
y2 + 18y + 81
4h2 – 28h + 49
1681
841
p6 – 49
896
9-4

49. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 9-5
(For help, go to Skills Handbook page 721.)
List all of the factors of each number.
9-5

50. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 9-5
Solutions
1. Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
2. Factors of 12: 1, 2, 3, 4, 6, 12
3. Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54
4. Factors of 15: 1, 3, 5, 15
5. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
6. Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56
7. Factors of 64: 1, 2, 4, 8, 16, 32, 64
8. Factors of 96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
9-5

51. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 9-5
Factor x2 + 8x + 15.
Find the factors of 15. Identify the pair that has a sum of 8.
Factors of 15 Sum of Factors
1 and
3 and
x2 + 8x + 15 = (x + 3)(x + 5).
= x2 + 5x + 3x + 15
Check: x2 + 8x (x + 3)(x + 5)
= x2 + 8x + 15
9-5

52. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 9-5
Factor c2 – 9c + 20.
Since the middle term is negative, find the negative factors of 20.
Identify the pair that has a sum of –9.
Factors of 20 Sum of Factors
–1 and –20 –21
–2 and –10 –12
–4 and –5 –9
c2 – 9c + 20 = (c – 5)(c – 4)
9-5

53. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 9-5
a. Factor x2 + 13x – 48.
b. Factor n2 – 5n – 24.
Identify the pair of factors of –48 that has a sum of 13.
Identify the pair of factors of –24 that has a sum of –5.
Factors of –48 Sum of Factors
1 and –48 –47
48 and –
2 and –24 –22
24 and –
3 and –16 –13
16 and –
Factors of –24 Sum of Factors
1 and –24 –23
24 and –
2 and –12 –10
12 and –
3 and –8 –5
x2 + 13x – 48 = (x + 16)(x – 3)
n2 – 5n – 24 = (n + 3)(n – 8)
9-5

54. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 9-5
Factor d2 + 17dg – 60g2.
Factors of –60 Sum of Factors
1 and –60 –59
60 and –
2 and –30 –28
30 and –
3 and –20 –17
20 and –
Find the factors of –60.
Identify the pair that has a sum of 17.
d2 + 17dg – 60g2 = (d – 3g)(d + 20g)
9-5

55. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 9-5
pages 483–485  Exercises
1. 5
2. 9
3. 7
4. 6
5. (r + 3)(r + 1)
6. (n – 2)(n – 1)
7. (k + 3)(k + 2)
8. (y + 4)(y + 2)
9. (x – 1)(x – 1)
10. (p + 18)(p + 1)
11. (k – 14)(k – 2)
12. (w + 5)(w + 1)
13. (m – 1)(m – 8)
14. (d + 19)(d + 2)
15. (t – 7)(t – 6)
16. (q – 15)(q – 3)
17. 5
18. 6
19. 9
20. 6
21. (x + 4)(x – 1)
22. (q – 4)(q + 2)
23. (y + 5)(y – 4)
24. (h + 17)(h – 1)
25. (x – 16)(x + 2)
26. (d + 10)(d – 4)
27. (m + 2)(m – 15)
28. (p – 6)(p + 9)
29. (p + 3)(p – 18)
30. A
31. B
32. B
33. (t + 9v)(t – 2v)
34. (x + 7y)(x + 5y)
35. (p – 8q)(p – 2q)
9-5

56. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 9-5
42. a. Factors contain the same operation.
b. Factors contain opposite operations.
43. (k + 2)(k + 8)
44. (m – 2)(m + 12)
45. (n – 4)(n + 14)
46. (g + 12)(g + 8)
47. (x – 5)(x + 13)
48. (t + 3)(t + 25)
49. (x – 14)(x + 3)
50. (k + 21)(k + 2)
51. (m – 3)(m + 17)
52. (x + 25y)(x + 4y)
36. (m – 9n)(m + 6n)
37. (h + 17j)(h + j)
38. (x – 13y)(x + 3y)
39–41. Answers may vary.
Samples are given.
39. 18; (x – 6)(x + 3)
28; (x – 7)(x + 4)
10; (x – 5)(x + 2)
40. 12; (x + 4)(x – 3)
2; (x + 2)(x – 1)
20; (x + 5)(x – 4)
41. 7; (x + 4)(x + 3)
8; (x + 6)(x + 2)
13; (x + 12)(x + 1)
9-5

57. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 9-5
53. (t – 15)(t + 5)
54. (d – 16e)(d – 3e)
55. 4x2 + 12x + 5; (2x + 1)(2x + 5)
56. 6x2 + 13x + 6; (3x + 2)(2x + 3)
57. a. The signs of a and b
must be opposite.
b. Since the middle term is
negative, the number with
the larger absolute value
must be negative. Therefore,
a must be a negative integer.
58. a. The signs of a and b must
be opposite.
b. Since the middle term is positive, the
number with the larger absolute value
must be positive. Therefore, b is a
negative integer.
59. (x6 + 7)(x6 + 5)
60. (t4 + 8)(t4 – 3)
61. (r 3 – 16)(r 3 – 5)
62. (m5 + 17)(m5 + 1)
63. (x6 – 24)(x6 + 5)
64. (p3 – 4)(p3 + 18)
65. B
66. F
67. D
9-5

58. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 9-5
68. I
69. A
70. F
71. [2] Find a pair of factors
of –40 that has a sum
of –18: –20 and 2.
x2 – 18x – 40 =
(x – 20)(x + 2)
[1] correct explanation with
incorrect factoring OR
incorrect explanation
with correct factoring
72. x2 + 8x + 16
73. w 2 – 12w + 36
74. r 2 – 25
75. 4q q + 49
76. 64v 2 – 4
77. 9a2 – 54a + 81
78. 9a2 – 25
79. 36t t + 81
80. 4x2 – 64y 2
81. 6 weeks
82. 7, 35
83. a. 81 basic players,
48 deluxe players
b. $
84.
85.
86.
9-5

59. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 9-5
Factor each expression.
1. c2 + 6c x2 – 11x g2 – 2g – 24
4. y2 + y – m2 – 2mn + n2
(c + 3)(c + 3)
(x – 2)(x – 9)
(g – 6)(g + 4)
(y + 11)(y – 10)
(m – n)(m – n)
9-5

60. Factoring Trinomials of the Type ax2 + bx + c
ALGEBRA 1 LESSON 9-6
(For help, go to Lessons 9-2 and 9-5.)
Find the greatest common factor.
1. 12x2 + 6x 2. 28m2 – 35m v3 + 36v2 + 10
Factor each expression.
4. x2 + 5x y2 – 3y – t2 – 11t + 30
9-6

61. Factoring Trinomials of the Type ax2 + bx + c
ALGEBRA 1 LESSON 9-6
Solutions
1. 12x2 + 6x 12x2 = 2 • 2 • 3 • x • x; 6x = 2 • 3 • x; GCF = 2 • 3 • x = 6x
2. 28m2 – 35m m2 = 2 • 2 • 7 • m • m; 35m = 5 • 7 • m; 14 = 2 • 7; GCF = 7
3. 4v3 + 36v v3 = 2 • 2 • v • v • v; 36v2 = 2 • 2 • 3 • 3 • v • v; 10 = 2 • 5; GCF = 2
4. Factors of 4 with a sum of 5: 1 and 4 x2 + 5x + 4 = (x + 1)(x + 4)
5. Factors of –28 with a sum of –3: 4 and –7 y2 – 3y – 28 = (y + 4)(y – 7)
6. Factors of 30 with a sum of –11: –5 and –6 t2 – 11t + 30 = (t – 5)(t – 6)
9-6

62. Factoring Trinomials of the Type ax2 + bx + c
ALGEBRA 1 LESSON 9-6
Factor 20x2 + 17x + 3.
20x2 + 17x + 3
F O I L
1 • 20 1 • • 20 = 23 1 • 3
1 • • 20 = 61 3 • 1
factors of a
factors of c
2 • 10 2 • • 10 = 16 1 • 3
2 • • 10 = 32 3 • 1
4 • 5 4 • • 5 = 17 1 • 3
20x2 + 17x + 3 = (4x + 1)(5x + 3)
9-6

63. Factoring Trinomials of the Type ax2 + bx + c
ALGEBRA 1 LESSON 9-6
Factor 3n2 – 7n – 6.
3n2 –7n –6
(1)(3)     (1)(–6) + (1)(3) = –3 (1)(–6)
(1)(1) + (–6)(3) = –17 (–6)(1)
(1)(–3) + (2)(3) = 3 (2)(–3)
(1)(2) + (–3)(3) = –7 (–3)(2)
3n2 – 7n – 6 = (n – 3)(3n + 2)
9-6

64. Factoring Trinomials of the Type ax2 + bx + c
ALGEBRA 1 LESSON 9-6
Factor 18x2 + 33x – 30 completely.
18x2 + 33x – 30 = 3(6x2 + 11x – 10) Factor out the GCF.
Factor 6x2 + 11x – 10.
6x2 + 11x –10
(2)(3) (2)(–10) + (1)(3) = –17 (1)(–10)
(2)(1) + (–10)(3) = –28 (–10)(1)
(2)(–5) + (2)(3) = –4 (2)(–5)
(2)(2) + (–5)(3) = –11 (–5)(2)
(2)(–2) + (5)(3) = 11 (5)(–2)
6x2 + 11x – 10 = (2x + 5)(3x – 2)
18x2 + 33x – 30 = 3(2x + 5)(3x – 2)
Include the GCF in your final answer.
9-6

65. Factoring Trinomials of the Type ax2 + bx + c
ALGEBRA 1 LESSON 9-6
pages 487–489  Exercises
1. (2n + 1)(n + 7)
2. (7d + 1)(d + 7)
3. (11w – 3)(w – 1)
4. (3x – 2)(x – 5)
5. (3t + 11)(2t + 1)
6. (3d – 5)(d – 4)
7. (2m + 1)(8m + 9)
8. (p – 1)(15p – 11)
9. (2y + 1)(4y + 13)
10. (2y + 1)(y + 17)
11. (x – 3)(7x – 9)
12. (4x + 3)(2x + 3)
13. (2t – 3)(t + 1)
14. (4y + 1)(2y – 3)
15. (2q + 3)(q – 7)
16. (7x + 1)(x – 3)
17. (13p – 5)(p + 1)
18. (5k – 7)(k + 1)
19. (5w + 8)(2w – 1)
20. (4d + 5)(3d – 4)
21. (7n + 15)(2n – 1)
22. 8(3m – 1)(m – 1)
23. 7(3v – 7)(v – 1)
24. 2(3t + 4)(t + 3)
25. 5(5x + 3)(x – 1)
26. 11(p + 1)(p + 6)
27. 2(4v + 3)(3v – 1)
28. Answers may vary. Sample:
41; (4g + 1)(g + 10)
14; (4g + 10)(g + 1)
22; (2g + 1)(2g + 10)
29. Answers may vary. Sample:
18; (5m – 4)(3m + 6)
54; (5m – 2)(3m + 12)
117; (5m – 1)(3m + 24)
30. Answers may vary. Sample:
8; (7g – 4)(5g + 4)
559; (35g – 1)(g + 16)
46; (7g – 2)(5g + 8)
9-6

66. Factoring Trinomials of the Type ax2 + bx + c
ALGEBRA 1 LESSON 9-6
36. 28(m – 1)(m + 2)
37. 3(7h – 4)(h + 4)
38. (11n – 6)(5n – 2)
39. 2(6y – 1)(3y + 10)
40. (9w – 5)(7w – 6)
41. (9q – 1)(11q – 9)
42. 2
43. Answers may vary. Sample:
5x2 – 12x + 4 = (5x – 2)(x – 2);
9x2 – 12x + 3 = 3(3x – 1)(x – 1);
16x2 – 12x + 2 = 2(4x – 1)(2x – 1)
44. x(8x + 5)(7x + 1)
45. (7p – 3q)(7p + 12q)
31. a. (2x + 2)(x + 2); (x + 1)(2x + 4)
b. 2x2 + 6x + 4; 2x2 + 6x + 4; yes
c. Answers may vary. Sample:
Neither factoring is complete.
Each one has a common factor, 2.
32. Answers may vary. Sample: Factor
out the GCF, 2, first. Look at the
factors of 25 and 8 to find a
combination that will give you a sum
of –45. 2(25x2 – 45x + 8)
= 2(5x – 1)(5x – 8).
33. (9p + 4)(6p + 7)
34. 3(11r + 4)(2r + 1)
35. (7x – 2)(2x – 7)
9-6

67. Factoring Trinomials of the Type ax2 + bx + c
ALGEBRA 1 LESSON 9-6
46. 54h(2g – 1)(g – 1)
47. a. –2 and –3
b. (x + 2)(x + 3)
c. Answers may vary. Sample:
Each x-intercept is the
opposite of the last term in
a binomial factor.
48. D
49. G
50. D
51. H
52. D
53. [2] 3x2 + 40x – 75 = (3x – 5)(x + 15)
[1] one computational error OR no work shown
54. (y + 7)(y + 1)
55. (t – 4)(t – 3)
56. (p – 5)(p + 4)
57. (m – 3)(m – 12)
58. (k + 18)(k – 2)
59. (g + 9)(g + 8)
60. (h – 16)(h + 3)
61. (x – 15)(x + 2)
62. (d – 4)(d – 14)
,801
,409
9-6

68. Factoring Trinomials of the Type ax2 + bx + c
ALGEBRA 1 LESSON 9-6
66. 38,809
69. 39,996
, 4, 16
72. – , –3, –27
, , 3
, 5, or
, , 2
, 8, 0.32
77.
78.
79.
80. 1 2 1 9 1 81 1 3 5
100 1 20 1
1250 1 10 1 2
9-6

69. Factoring Trinomials of the Type ax2 + bx + c
ALGEBRA 1 LESSON 9-6
Factor each expression.
1. 3x2 – 14x + 11
2. 6t2 + 13t – 63
3. 9y2 – 48y – 36
(x – 1)(3x – 11)
(2t + 9)(3t – 7)
3(3y + 2)(y – 6)
9-6

70. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
(For help, go to Lessons 8–4 and 9-4.)
Simplify each expression.
1. (3x)2 2. (5y)2 3. (15h2)2 4. (2ab2)2
Simplify each product.
5. (c – 6)(c + 6) 6. (p – 11)(p – 11) 7. (4d + 7)(4d + 7)
9-7

71. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
Solutions
1. (3x)2 = 32 • x2 = 9x2 2. (5y)2 = 52 • y2 = 25y2
3. (15h2)2 = 152 • (h2)2 = 225h4 4. (2ab2)2 = 22 • a2 • (b2)2 = 4a2b4
5. (c – 6)(c + 6) is the difference of squares. (c – 6)(c + 6) = c2 – 62 = c2 – 36
6. (p – 11)(p – 11) is the square of a binomial. (p – 11)2 = p2 – 2p(11) = p2 – 22p + 121
7. (4d + 7)(4d + 7) is the square of a binomial. (4d + 7)2 = (4d)2 + 2(4d)(7) + 72 = 16d2 + 56d + 49
9-7

72. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
Factor m2 – 6m + 9.
m2 – 6m + 9 = m • m – 6m + 3 • 3 Rewrite first and last terms.
= m • m – 2(m • 3) + 3 • 3 Does the middle term equal 2ab? 6m = 2(m • 3)
= (m – 3)2 Write the factors as the square of a binomial.
9-7

73. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
The area of a square is (16h2 + 40h + 25) in.2. Find the length of a side.
16h2 + 40h + 25 = (4h)2 + 40h + 52 Write 16h2 as (4h)2 and 25 as 52.
= (4h)2 + 2(4h)(5) + 52 Does the middle term equal 2ab? 40h = 2(4h)(5)
= (4h + 5)2 Write the factors as the square of a binomial.
The side of the square has a length of (4h + 5) in.
9-7

74. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
Factor a2 – 16.
a2 – 16 = a2 – 42 Rewrite 16 as 42.
= (a + 4)(a – 4) Factor.
Check: Use FOIL to multiply.
(a + 4)(a – 4)
a2 – 4a + 4a – 16
a2 – 16
9-7

75. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
Factor 9b2 – 25.
9b2 – 225 = (3b)2 – 52 Rewrite 9b2 as (3b)2 and 25 as 52.
= (3b + 5)(3b – 5) Factor.
9-7

76. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
Factor 5x2 – 80.
5x2 – 80 = 5(x2 – 16) Factor out the GCF of 5.
= 5(x + 4)(x – 4) Factor (x2 – 16).
Check: Use FOIL to multiply the binomials. Then multiply by the GCF.
5(x + 4)(x – 4)
5(x2 – 16)
5x2 – 80
9-7

77. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
pages 493–495  Exercises
1. (c + 5)2
2. (x – 1)2
3. (h + 6)2
4. (m – 12)2
5. (k – 8)2
6. (t – 7)2
7. (2m + 5)
8. (7d + 2)
9. (5g – 4)
10. (5g – 3)2
11. (8r – 9)2
12. (10v – 11)2
13. (x + 2)(x – 2)
14. (y + 9)(y – 9)
15. (k + 14)(k – 14)
16. (r + 12)(r – 12)
17. (h + 10)(h – 10)
18. (m + 15)(m – 15)
19. (w + 16)(w – 16)
20. (x + 20)(x – 20)
21. (y + 30)(y – 30)
22. (5q + 3)(5q – 3)
23. (7y + 2)(7y – 2)
24. (3c + 8)(3c – 8)
25. (2m + 9)(2m – 9)
26. (4k + 7)(4k – 7)
27. (12p + 1)(12p – 1)
28. (9v + 10)(9v – 10)
29. (20n + 11)(20n – 11)
30. (5w + 14)(5w – 14)
31. 3(m + 2)(m – 2)
32. 5(k + 7)(k – 7)
33. 3(x + 8)2
34. 2(t – 9)2
35. 6r (r + 5)(r – 5)
9-7

78. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
44. a. Answers may vary.
Sample: 4x2 + 24x + 36
b. because (2x)2 = 4x2,
2(2x • 6) = 24x, and
62 = 36
45. 25(2v + w)(2v – w)
46. 4(2p – 3q)2
47. 7(2c + 5d)2
m m –
49. x +
50. 16(2g – 3h)2
p – 2
36. 7(h – 4)2
37. Answers may vary. Sample: Rewrite the
first and last terms as a square. Check to
see if the middle term is 2ab. Factor as a
square binomial: 4x2 + 12x + 9 =
(2x)2 + 12x + 32 = (2x)2 + 2(2x)(3) + 32 =
(2x + 3)2; 9x2 – 30x + 25 = (3x)2 – 30x + 52
= (3x)2 – 2(3x)(5) + 52 = (3x – 5)2.
38. 4x2 – 121 is the difference of two squares.
So the answer should be (2x + 11)(2x – 11).
39. 11, 9
40. 13, 7
41. 15, 5
42. 13, 9
43. 16, 14 1 2 1 3 1 2 1 3 1 2 2 1 2 2
9-7

79. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
n n –
k + 3
54. a. 3.14n2 – 3.14m2 =
3.14(n + m)(n – m)
b in.2
55. a. 4(x + 5)(x – 5)
b. 4(x + 5)(x – 5)
c. The polynomial has a
GCF that has two
identical factors.
d. 3(x + 5)(x – 5); no,
because 3 does not
have a pair of
56. (8r 3 – 9)2
57. (p3 + 20q)2
58. (6m2 + 7)2
59. (9p5 + 11)2
60. 3(6m3 – 7)(6m3 + 7)
61. (x10 – 2y5)2
62. 4(8g 2 – 5h3)(8g 2 + 5h3)
63. 5(3x2 – 2y)2
64. 37(g4 + h4)(g 2 + h2)(g + h)(g – h)
65. a. t – 3; 4
b. (t + 1)(t – 7) 1 3 1 5 1 3 1 5 1 5 2
9-7

80. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
75. (2t + 1)(2t + 7)
76. (5w + 1)(w – 9)
77. (3t + 8)(2t + 1)
78. (7m – 9)(3m + 1)
79. (7x – 9)(2x + 1)
80. (2y + 11)(2y + 5)
81. (3k – 2)(4k + 1)
; 3072; 12,288; 3 • 4n-1
83. 29; 37; 45; –11 + 8n
84. –11; –20; –29; 34 – 9n
; 0.002; ; 2000 • n
86. –32; 64; –128; (–2)n
66. a. (4 + 9n2)(2 + 3n)(2 – 3n)
b. They are squares of
square terms.
c. Answers may vary.
Sample: 16x4 – 1
67. 9
68. –12
69. 30
70. 5
71. 12
73. (2d + 1)(d + 5)
74. (2x – 3)(x – 4) 1 10
9-7

81. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
; 777.6; ; 0.1(6)n–1
; ; ; 10 •
; 8; 9 ; – n
90. 8; 32; 128; • 4n–1 or • 4n–3
91. a. y = 11.4x
b. 93
c. 79
n–1 1 2 3 32
125 64
625
128
3125 5
; 777.6; ; 0.1(6)n–1
; ; ; 10 •
; 8; 9 ; – n
90. 8; 32; 128; • 4n–1 or • 4n–3
91. a. y = 11.4x
b. 93
c. 79
n–1 1 2 3 32
125 64
625
128
3125 5
; 777.6; ; 0.1(6)n–1
; ; ; 10 •
; 8; 9 ; – n
90. 8; 32; 128; • 4n–1 or • 4n–3
91. a. y = 11.4x
b. 93
c. 79
n–1 1 2 3 32
125 64
625
128
3125 5
9-7

82. Factoring Special Cases
ALGEBRA 1 LESSON 9-7
Factor each expression.
1. y2 – 18y a2 – 24a + 16
3. p2 – x2 – 225
5. 5m2 – c2 + 20c + 50
(y – 9)2
(3a – 4)2
(p + 13)(p – 13)
(6x + 15)(6x – 15)
5(m + 3)(m – 3)
2(c + 5)2
9-7

83. Find the GCF of the terms of each polynomial.
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
(For help, go to Lessons 9-2 and 9-3.)
Find the GCF of the terms of each polynomial.
1. 6y2 + 12y – r3 + 15r2 + 21r
3. 30h3 – 25h2 – 40h 4. 16m3 – 12m2 – 36m
Find each product.
5. (v + 3)(v2 + 5) 6. (2q2 – 4)(q – 5)
7. (2t – 5)(3t + 4) 8. (4x – 1)(x2 + 2x + 3)
9-8

84. 12y = 2 • 2 • 3 • y; 4 = 2 • 2; 15r2 = 3 • 5 • r • r; 21r = 3 • 7 • r;
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
Solutions
1. 6y2 + 12y – r3 + 15r2 + 21r
6y2 = 2 • 3 • y • y; 9r3 = 3 • 3 • r • r • r;
12y = 2 • 2 • 3 • y; 4 = 2 • 2; 15r2 = 3 • 5 • r • r; 21r = 3 • 7 • r;
GCF = 2 GCF = 3r
3. 30h3 – 25h2 – 40h 4. 16m3 – 12m2 – 36m
30h3 = 2 • 3 • 5 • h • h • h; 16m3 = 2 • 2 • 2 • 2 • m • m • m;
25h2 = 5 • 5 • h • h; 12m2 = 2 • 2 • 3 • m • m;
40h = 2 • 2 • 2 • 5 • h; 36m = 2 • 2 • 3 • 3 • m;
GCF = 5h GCF = 2 • 2 • m = 4m
5. (v + 3)(v2 + 5) = (v)(v2) + (v)(5) + (3)(v2) + (3)(5)
= v3 + 5v + 3v2 + 15
= v3 + 3v2 + 5v + 15
9-8

85. Solutions (continued)
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
Solutions (continued)
6. (2q2 – 4)(q – 5)
= (2q2)(q) + (2q2)(–5) + (–4)(q) + (–4)(–5)
= 2q3 – 10q2 – 4q + 20
7. (2t – 5)(3t + 4)
= (2t)(3t) + (2t)(4) + (–5)(3t) + (–5)(4)
= 6t2 + 8t – 15t – 20
= 6t2 – 7t – 20
8. (4x – 1)(x2 + 2x + 3) = (4x)(x2) + (4x)(2x) + (4x)(3)
+ (–1)(x2) + (–1)(2x) + (–1)(3)
= 4x3 + 8x2 + 12x – x2 – 2x – 3
= 4x3 + (8 – 1)x2 + (12 – 2)x – 3
= 4x3 + 7x2 + 10x – 3
9-8

86. = (2x + 1)(3x2 – 2) Factor out (2x + 1).
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
Factor 6x3 + 3x2 – 4x – 2.
6x3 + 3x2 – 4x – 2 = 3x2(2x + 1) – 2(2x + 1) Factor the GCF from each group of two terms.
= (2x + 1)(3x2 – 2) Factor out (2x + 1).
= 6x3 – 4x + 3x2 – 2 Use FOIL.
Check: 6x3 + 3x2 – 4x – (2x + 1)(3x2 – 2)
= 6x3 + 3x2 – 4x – 2 Write in standard form.
9-8

87. 8t4 + 12t3 + 16t + 24 = 4(2t4 + 3t3 + 4t + 6) Factor out the GCF, 4.
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
Factor 8t4 + 12t3 + 16t + 24.
8t4 + 12t3 + 16t + 24 = 4(2t4 + 3t3 + 4t + 6) Factor out the GCF, 4.
= 4[t3(2t + 3) + 2(2t + 3)] Factor by grouping.
= 4(2t + 3)(t3 + 2) Factor again.
9-8

88. Step 1: 24h2 + 10h – 6 = 2(12h2 + 5h – 3) Factor out the GCF, 2.
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
Factor 24h2 + 10h – 6.
Step 1: 24h2 + 10h – 6 = 2(12h2 + 5h – 3)   Factor out the GCF, 2.
Step 2: 12 • –3 = –36 Find the product ac.
Step 3: Factors Sum
–2(18) = –36 – = 16
–3(12) = –36 – = 9
–4(9) = –36 –4 + 9 = 5
Find two factors of ac that have a sum b. Use mental math to determine a good place to start.
Step 4:  12h2 – 4h + 9h – 3 Rewrite the trinomial.
Step 5:   4h(3h – 1) + 3(3h – 1) Factor by grouping.
  (4h + 3)(3h – 1) Factor again.
24h2 + 10h – 6 = 2(4h + 3)(3h – 1) Include the GCF in your final answer.
9-8

89. Step 1: 3x(12x2 + 17x + 6) Factor out the GCF, 3x.
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
A rectangular prism has a volume of 36x3 + 51x2 + 18x. Factor to find the possible expressions for the length, width, and height of the prism.
Factor 36x3 + 51x2 + 18x.
Step 1: 3x(12x2 + 17x + 6) Factor out the GCF, 3x.
Step 2: 12 • 6 = 72 Find the product ac.
Step 3:  Factors     Sum
4 • = 22
6 • = 18
8 • = 17
Find two factors of ac that have sum b. Use mental math to determine a good place to start.
9-8

90. Step 4: 3x(12x2 + 8x + 9x + 6) Rewrite the trinomial.
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
(continued)
Step 4: 3x(12x2 + 8x + 9x + 6) Rewrite the trinomial.
Step 5: 3x[4x(3x + 2) + 3(3x + 2)] Factor by grouping.
3x(4x + 3)(3x + 2) Factor again.
The possible dimensions of the prism are 3x, (4x + 3), and (3x + 2).
9-8

91. Factoring by Grouping pages 499–501 Exercises
ALGEBRA 1 LESSON 9-8
pages 499–501  Exercises
1. 2m2; 3
2. 5p2; 2
3. 2z2; –5
4. 3n2; 1
5. (2n2 + 1)(3n + 4)
6. (7t 2 + 8)(2t + 3)
7. (3t + 1)(3t – 1)(3t + 5)
8. (y2 + 1)(13y – 8)
9. (5x2 + 1)(9x + 4)
10. (2w2 – 3)(5w + 8)
11. 2(2v2 + 1)(3v – 8)
12. q(q 2 + 4)(7q – 4)
13. 2(m2 + 2)(10m – 9)
14. 2x(x + 1)(x – 1)(3x + 2)
15. 2(2y2 + 5)(3y – 5)
16. 3(c2 + 2)(3c – 4)
17. (6p + 5)(2p + 1)
18. (4t + 3)2
19. (6n – 1)(3n + 10)
20. (3w – 5)(3w – 4)
21. 2(6m – 1)(2m + 1)
22. (12v – 7)(3v + 1)
23. (3x – 2)(2x + 5)
24. (4v – 1)(5v – 9)
25. (7q + 2)(9q – 10)
26. m, (3m + 1), and
(m + 2)
27. 5k, (k + 2), and (k + 4)
28. 7h(h – 6)(h + 1)
29. 2(10t 2 – 11)(3t – 10)
30. 8(d 2 + 3)(d + 2)
31. 4(3x – 7y)(x + 2y)
32. 9r (3r – 1)(2r – 1)
33. 10(5k2 + 6)(3k + 7)
34. a. (7x2 + 9)(4x – 1)
b. (4x – 1)(7x2 + 9)
c. Answers may vary. Sample: The factorings are equivalent; but the factors may appear in a different order.
9-8

92. 40. Answers may vary. Sample: 30x2 + 36x + 40x + 48; 2(3x + 4)(5x + 6)
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
35. (7w 2 – 4)(2w + 7)
36. (2m2 – 1)(m – 16)
37. 2(2t 2 + 3)(11t – 1)
38. (x2 – 2)(25x – 1)
39. 2w, (6w + 5), and (7w + 1)
40. Answers may vary. Sample:
30x2 + 36x + 40x + 48;
2(3x + 4)(5x + 6)
41. Answers may vary. Sample:
Split the expression into two
groups. Remove the GCF
from each group, and then
factor again.
42. (6m3 – 7n2)(5m2 + 4n)
43. (x2 + y)(p + q5)
44. (h + 2)(h – 2)(h + 11)
45. (w 2 + 3)(w 2 – 3)(w + 1)(w – 1)
46. a x(x + 3)2
b. x + 3
47. ( )( ); (9)(7)
48. ( ) ( ); (21)(3)
49. Answers may vary. Samples are given.
a. length = 2x + 4; width = x; height = x + 4
b. 2x3 + 12x2 + 16x
50. C
51. H
52. [2] 9a4 – 54a3 – 2a + 12 =
9a3(a – 6) – 2(a – 6) = (9a3 – 2)(a – 6)
[1] appropriate methods with one
computational error
9-8

93. solutions on the line y = –4x+ 25 84. (7, 2)
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
58. (m + 8)(m – 8)
59. 4(g + 5)2
60. (2d + 5)(2d – 5)
61. 5(n + 3)(n – 3)
62. (5q + 4)2
63. b4
64. x2
65. t 15
66. c35d 7
67. 8y3
68. 1
69.
70. 81w8v12
 1021 1
x11
72. 9  1012
 10–11
 1036
 105
 10–5
 1024
 104
79. (1, 5)
80. (4, –8)
81. (0.5, 5)
82. (2, –30)
83. infinitely many
solutions on the line y = –4x+ 25
84. (7, 2)
53. [4] 96x3 + 48x2 + 6x =
6x(16x2 + 8x + 1) =
6x(4x + 1)2. Side of
square equals 4x + 1.
Perimeter = 4(4x + 1) =
16x + 4.
[3] appropriate methods,
but with one
computational error
[2] found factors of polynomial,
but did not find perimeter
[1] correct answer, without
work shown
54. (k + 7)2
55. (r + 3)2
56. (y – 8)2
57. 2(t + 3)2
9-8

94. Factor each expression. 1. 10p3 – 25p2 + 4p – 10
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
Factor each expression.
1. 10p3 – 25p2 + 4p – 10
2. 36x4 – 48x3 + 9x2 – 12x
3. 16a3 – 24a2 + 12a – 18
(5p2 + 2)(2p – 5)
3x(4x2 + 1)(3x – 4)
2(4a2 + 3)(2a – 3)
9-8

95. Polynomials and Factoring
ALGEBRA 1 CHAPTER 9
1. –2y2 – 3y + 5; quadratic trinomial
2. –2v3 – 6v2 – 17v; cubic trinomial
3. 8x4 – 6x2 – 10; fourth degree trinomial
4. –3k5 – k2; fifth degree binomial
5. 11x2 – 3x + 7
6. 15a2 – 17a – 7
7. –11m2 + 4m + 2
8. –7c3 – 9c2 + 9c
9. Answers may vary. Sample: p6 + p2 + 1
10. x2 + 5x + 6; (x + 3)(x + 2)
11. –8b3 + 24b2 + 56b
12. –5t 3 – t 2
13. 9q4 – 3q2 + 12q
14. 2c6 + 8c4
15. x2 + 7x + 6
16. d2 + d – 12
17. 2h2 – 9h + 4
18. 6m2 + m – 35
19. 2p3 – p2 – 6p + 8
20. 6a3 – 14a2 – 43a + 12
21. 21x3 + 29x2 – 7x + 5
22. 3x2
23. t
24. 3a5
9-A

96. Polynomials and Factoring
ALGEBRA 1 CHAPTER 9
25. m2
26. Multiply each term of the first polynomial
by each term of the second polynomial.
Then combine like terms. Example:
(x + 3)(x2 – x + 1) = x3 – x2 + x + 3x2 – 3x + 3
= x3 + 2x2 – 2x + 3
27. x(3x + 5); (3x2 + 5x) m2
28. w(w – 2)(4w + 3); (4w 3 – 5w 2 – 6w) in.3
29. 4x2 – 3x
30. 4x2 + x
31. (w – 7)(w + 2)
32. (g + 5)2
33. (3k + 4)2
34. (n + 10)(n – 10)
35. (y – 2)2
36. (2x + 7)(2x – 7)
37. (2p + 81)(2p + 1)
38. 13(c + 2)(c – 2)
39. 14
40. 4
41. 25
42. 36
43. 3x3; – 4
44. 4n2; –1
45. (3n2 + 1)(4n + 5)
9-A

97. Polynomials and Factoring
ALGEBRA 1 CHAPTER 9
46. 2(2x – 3)(x – 1)
47. (x2 + 5)(x – 5)
48. (3r 2 – 2)(2r – 3)
49. (2y + 1)(2y – 1)(3y + 7)
50. (n2 – 2)(3n – 4)
51. Answers may vary. Sample:
13, (x + 10)(x + 3);
11, (x + 5)(x + 6);
31, (x + 1)(x + 30)
9-A