# Factoring complex trinomials

• Apr 26th 2012, 06:04 AM
Capnzims
Factoring complex trinomials
6x^2+6^x-12

I get an answer that works of (6x-6)(x+2) but the book says
(2x+4)(3x-3). What do I have to do to match the book or what did I do wrong?
• Apr 26th 2012, 06:19 AM
princeps
Re: Factoring complex trinomials
Quote:

Originally Posted by Capnzims
6x^2+6^x-12

I get an answer that works of (6x-6)(x+2) but the book says
(2x+4)(3x-3). What do I have to do to match the book or what did I do wrong?

$\displaystyle 6x^2+6x-12=6(x^2+x-2)=2\cdot 3 \cdot(x+2)(x-1)=(2x+4)(3x-3)$
• Apr 26th 2012, 06:23 AM
Capnzims
Re: Factoring complex trinomials
So factor the 6 outside?
• Apr 26th 2012, 06:27 AM
princeps
Re: Factoring complex trinomials
yes
• Apr 26th 2012, 06:45 AM
Capnzims
Re: Factoring complex trinomials
Apply the 2 to the first () and the 3 to the next ()? Can I always factor like that first before working the problem on trinomials?
• Apr 26th 2012, 06:55 AM
princeps
Re: Factoring complex trinomials
yes
• Apr 26th 2012, 07:06 AM
masters
Re: Factoring complex trinomials
Quote:

Originally Posted by Capnzims
6x^2+6^x-12

I get an answer that works of (6x-6)(x+2) but the book says
(2x+4)(3x-3). What do I have to do to match the book or what did I do wrong?

Hi Capnzims,

When asked to factor a polynomial that has a common factor over all terms, you should always factor that out first. Then, factor the remaining polynomial, if possible.

$\displaystyle \displaystyle {6x^2+6x-12=6(x^2+x-2)=6(x+2)(x-1)}$

You wouldn't use $\displaystyle \displaystyle {(2x+4)(3x-3)}$ because there are common factors in each binomial.
• Apr 26th 2012, 05:38 PM
HallsofIvy
Re: Factoring complex trinomials
Now, do you understand that 6(x- 1)(x+ 2)= (6x- 6)(x+2)= (2x+ 4)(3x- 3)?

If you multiply the "6" into x- 1, you get (6x- 6)(x+ 2). If, instead, you factor 6 into (2)(3), multiply the "3" into (x- 1) and the "2" into x+ 2, you get (2x+ 4)(3x- 3).