1. 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)
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2. 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)
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3. 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
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4. 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.
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