1. ## 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?

2. ## Re: Factoring complex trinomials

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?
$6x^2+6x-12=6(x^2+x-2)=2\cdot 3 \cdot(x+2)(x-1)=(2x+4)(3x-3)$

3. ## Re: Factoring complex trinomials

So factor the 6 outside?

yes

5. ## 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?

yes

7. ## Re: Factoring complex trinomials

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.

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

You wouldn't use $\displaystyle {(2x+4)(3x-3)}$ because there are common factors in each binomial.

8. ## 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).