How do you factor # 6x^3+34x^2-12x#?

1 Answer
Oct 9, 2015

Answer:

#2x(3x-1)(1x+6)#

Explanation:

Given
#color(white)("XXX")6x^3+34x^2-12x#

First extract the obvious common factor of #(2x)# from each term:
#color(white)("XXX")(2x)(3x^2+17x-6)#

In the hopes of finding integer coefficient factors of #(3x^2+17x-6)#
we consider the integer factors of #3# and of #(-6)#
looking for a combination that will give us a sum of products #=17#

There are only a few possibilities. The diagram below might help understand the process:
enter image source here
The "working" combination is #(3x-1)(1x+6)#

Which gives us the complete factorization:
#color(white)("XXX")6x^3+34x^2-12x#
#color(white)("XXXXXXXXXXX")=(2x)(3x-1)(1x+6)#