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

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

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

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 2 + 28q + 49 76. 64v 2 – 4 77. 9a2 – 54a + 81 78. 9a2 – 25 79. 36t 2 + 108t + 81 80. 4x2 – 64y 2 81. 6 weeks 82. 7, 35 83. a. 81 basic players, 48 deluxe players b. $9719.19 84. 85. 86. 9-5