1. 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)
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2. 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
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3. 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
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4. 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
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5. 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