1. 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

2. 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

3. 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