﻿ STILL MORE FACTORING PROBLEMS

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

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)