How do you factor #3(x-4)^2+16(x-4) -12#?

2 Answers
Nov 28, 2015

Answer:

Factor as a quadratic in #(x-4)# then simplify at the last stage to find:

#3(x-4)^2+16(x-4)-12 = (x+2)(3x-14)#

Explanation:

One way is to leave the #(x-4)#'s intact until the end and look for a pair of factors of #3 xx 12 = 36#, whose difference is #16#. The pair #2#, #18# works, so use that to split the middle term and factor by grouping:

#3(x-4)^2+16(x-4)-12#

#=3(x-4)^2-2(x-4)+18(x-4)-12#

#=(3(x-4)^2-2(x-4))+(18(x-4)-12)#

#=(x-4)(3(x-4)-2)+6(3(x-4)-2)#

#=((x-4)+6)(3(x-4)-2)#

#=(x+2)(3x-14)#

Another way of expressing this solution is to substitute #t = x-4# and proceed as follows:

#3(x-4)^2+16(x-4)-12#

#=3t^2+16t-12#

#=3t^2-2t+18t-12#

#=(3t^2-2t)+(18t-12)#

#=t(3t-2)+6(3t-2)#

#=(t+6)(3t-2)#

#=((x-4)+6)(3(x-4)-2)#

#=(x+2)(3x-14)#

Nov 29, 2015

Answer:

Alternatively, multiply out the original quadratic and simplify before factoring to find:

#3(x-4)^2+16(x-4)-12 = (x+2)(3x-14)#

Explanation:

#3(x-4)^2+16(x-4)-12#

#=3(x^2-8x+16)+16(x-4)-12#

#=3x^2-24x+48+16x-64-12#

#=3x^2-8x-28#

Look for a pair of factors of #3 xx 28 = 84# with difference #8#.

The pair #6#, #14# works.

So:

#3x^2-8x-28#

#=3x^2-14x+6x-28#

#=(3x^2-14x)+(6x-28)#

#=x(3x-14)+2(3x-14)#

#=(x+2)(3x-14)#