Unit 4 Review Solutions. 1. 12a 5 – 32a 2 4a 2 (3a 3 – 8) 2. a 2 b 2 + ab ab(ab + 1) 3. x(x – 2) + y(2 – x) x(x – 2) – y(x – 2) (x – 2)(x – y)

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Unit 4 Review Solutions

1. 12a 5 – 32a 2 4a 2 (3a 3 – 8) 2. a 2 b 2 + ab ab(ab + 1) 3. x(x – 2) + y(2 – x) x(x – 2) – y(x – 2) (x – 2)(x – y)

4. 2x(7 + b) – y(b + 7) (b + 7)(2x – y) 5. 2ra + a 2 – 2r – a a(2r + a) – 1(2r + a) (2r + a)(a – 1)

6. z 2 – 4z – 45 (z + 5)(z – 9) 7. b b + 40 (b + 8)(b + 5) 8. y 2 – y – 72 (y – 9)(y + 8)

9. 5z 2 – 15z – 140 5(z 2 – 3z – 28) 5(z – 7)(z + 4) 10. x 2 – 5xy + 6y 2 (x – 3y)(x – 2y)

11. 6y 2 – 5y + 1 (3y – 1)(2y – 1) 12. 2x 3 – 3x 2 – 5x x(2x 2 – 3x – 5) x(2x – 5)(x + 1) 13. 6a a + 21 (3a + 7)(2a + 3)

14. 15a a 3 + 7a 2 a 2 (15a a + 7) a 2 (5a + 7)(3a + 1) 15. 9x 2 – 16y 2 (3x – 4y)(3x + 4y) y 2 – 48yz + 9z 2 (8y – 3z)(8y – 3z) (8y – 3z) 2

17. 12n 2 – 48 12(n 2 – 4) 12(n – 2)(n + 2) 18. a 4 – 16 (a 2 – 4)(a 2 + 4) (a – 2)(a + 2)(a 2 + 4)

19. (z + 8)(z – 9) = 0 z + 8 = 0 z – 9 = 0 z = -8z = a 2 – 5a = 24 a 2 – 5a – 24 = 0 (a – 8)(a + 3) = 0 a – 8 = 0 a + 3 = 0 a = 8a = - 3

21. y(y + 8) = -15 y 2 + 8y = -15 y 2 + 8y + 15 = 0 (y + 5)(y + 3) = 0 y + 5 = 0 y + 3 = 0 y = - 5y = - 3

22. The sum of two numbers is eight. The sum of the squares of the two numbers is thirty- four. Find the two numbers. Let: x = the 1 st number 8 – x = the 2 nd number x 2 + (8 – x) 2 = 34 x 2 + (64 – 16x + x 2 ) = 34 2x 2 – 16x + 64 = 34 2x 2 – 16x + 30 = 0 2(x 2 – 8x + 15) = 0 2(x – 5)(x – 3) = 0 x – 5 = 0 x – 3 = 0 x = 5x = 3 The two numbers are 5 and 3