1. ## Factoring polynomials

I'm in a college algebra class and we're getting started with factoring. I'm going through most of them just fine but I'm stuck on this:

2(x+1)(x-3)^2 - 3(x+1)^2(x-3)

I'm guessing I multiply everything out, and I did, and got to this point:

-x^2(x+7) + 3(7x+9)

Is that it? Is it even right? I'd appreciate some help Thanks!!

2. I think you want to factor, not to expand the polynomial.

$\displaystyle 2(x+1)(x-3)^2-3(x+1)^2(x-3)=(x+1)(x-3)[2(x-3)-3(x+1)]=$

$\displaystyle =(x+1)(x-3)(-x-9)=-(x+1)(x-3)(x+9)$

3. Originally Posted by shenaniganny
I'm in a college algebra class and we're getting started with factoring. I'm going through most of them just fine but I'm stuck on this:

2(x+1)(x-3)^2 - 3(x+1)^2(x-3)

I'm guessing I multiply everything out, and I did, and got to this point:

-x^2(x+7) + 3(7x+9)

Is that it? Is it even right? I'd appreciate some help Thanks!!
Factor out the common factors:

$\displaystyle 2(x+1)(x-3)^2 - 3(x+1)^2(x-3) = 2(x+1)(x-3)(x-3) - 3(x+1)(x+1)(x-3)$

$\displaystyle {\color{white}2(x+1)(x-3)^2 - 3(x+1)^2(x-3)} = (x+1)(x-3)\left( 2(x-3) - 3(x+1) \right)$

$\displaystyle {\color{white}2(x+1)(x-3)^2 - 3(x+1)^2(x-3)} =(x+1)(x-3)\left( -x-9)\right)$

$\displaystyle {\color{white}2(x+1)(x-3)^2 - 3(x+1)^2(x-3)} =- (x+1)(x-3)\left( x+9)\right)$

4. heh seems really obvious now, thank you both, I appreciate it!