Factor using AC method

Question

Factor using the AC method.

 

−6ab² + 63ab − 135a

 

4k² − 32k + 48

Answer 1

-6ab²+63ab-135a = -3a(2b²-21b+45)

 

                               (2)(45) = 90

 

                               So, we need two numbers whose product is 90                                    and whose sum is -21.  We see that the numbers                                are -6 and -15.  Rewrite -21b as -6b-15b and                                      factor by grouping.

 

                         = -3a(2b²-6b-15b+45)

 

                         = -3a[2b(b-3) -15(b-3)]

 

                         = -3a(2b-15)(b-3)

 ------------------------------------------------------------------------------------

 4k²-32k+48 = 4(k²-8k+12)

 

                        (1)(12) = 12

 

                        Find two numbers whose product is 12 and whose sum                         is -8.  The nombers are -2 and -6.  Rewrite -8k as

                        -2k-6k and factor by grouping.  

 

                   = 4(k²-2k-6k+12)

 

                   = 4[k(k-2)-6(k-2)]

 

                   = 4(k-6)(k-2)