Factoring polynomials

**shenaniganny** · August 30th 2009, 11:08 PM

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!!

**red_dog** · August 30th 2009, 11:23 PM

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

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

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

**earboth** · August 30th 2009, 11:27 PM

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:

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

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

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

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

**shenaniganny** · August 30th 2009, 11:29 PM

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

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