STILL MORE FACTORING PROBLEMS

Courtesy of Harold Hiken

Factor each of the following expressions either by inspection or by using the grouping method discussed in your text. If a greatest common factor can be removed first, factor it out. If the problem cannot be factored, state so.

1) 6a² – 7ab – 5b² 5) 25a² + 25ab + 6b² 9) 10y² + 11yz – 18z²

2) 6x² – 5xy + y² 6) 12p² + 7pq – 12q² 10) 6m⁶n + 7m⁵n² + 2m⁴n³

3) 6m² – 5mn – 6n² 7) 4k⁴ – 2k³w – 6k²w² 11) 12k³q⁴ – 4k²q⁵ – kq⁶

4) 6a² – 47ab – 63b² 8) 4a⁴ – a³b – 3a²b² 12) 18z³y – 3z²y² – 105zy³

13) 18x²(y – 3)² + 15x(y – 3)² – 75(y – 3)² 16) 8(x + y)² + 14(x + y) – 15

14) 25q²(m + 1)³ – 5q(m + 1)³ – 2(m + 1)³ 17) 24(x – 1)² – 14(x – 1) – 3

15) 10(a – b)² – 11(a – b) – 6 18) 14(2 – x)² – 15(2 – x) – 11

Answers:

1) (3a – 5b)(2a + b)

2) (3x – y)(2x – y)

3) (3m + 2n)(2m – 3n)

4) (6a + 7b)(a – 9b)

5) (5a + 6b)(5a – b)

6) (4p – 3q)(3p + 4q)

7) 2k²(2k – 3w)(k + w)

8) a²(4a + 3b)(a – b)

9) (10y – 9z)(y + 2z)

10) m⁴n(3m + 2n)(2m + n)

11) kq⁴(6k + q)(2k – q)

12) 3zy(3z + 7y)(2z – 5y)

13) 3(y – 3)²(3x – 5)(2x + 5)

14) (m + 1)³(5q + 1)(5q – 2)

15) (5(a – b) + 2)(2(a – b) – 3)

16) (4(x + y) – 3)(2(x + y) + 5)

17) (6(x – 1) + 1)(4(x – 1) – 3) = (6x – 5)(4x – 7)

18) (7(2 – x) – 11)(2(2 – x) + 1) = (3 – 7x)(5 – 2x)