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)