# Factoring help, again...

• Nov 1st 2012, 06:19 PM
Nervous
Factoring help, again...
$-(x-2)(x^2+2x+5)$
$-(x^3+2x^2+5x-2x^2+4x-10)$
$-(x^3+x-10)$
$-x^3-x+10$

How do you factor this? I tried adding in a $0x^2$ and grouping, but that didn't work out. The other group can't factor out to $x-0$...
• Nov 1st 2012, 06:22 PM
chiro
Re: Factoring help, again...
Hey Nervous.

Try factoring the quadratic by getting the roots through the quadratic formula.
• Nov 1st 2012, 06:51 PM
Nervous
Re: Factoring help, again...
I thought quadratics were only second degree polynomials...?

EDIT: Ohhhhh I get it, you meant the first trinomial... OK...
• Nov 1st 2012, 06:52 PM
HallsofIvy
Re: Factoring help, again...
Why in the world would you first multiply and then factor? You start with $-(x- 2)(x^2+ 2x+ 5)$ so obviously, one factor is x- 2. In order to factor $x^2+ 2x+ 5$ into factors with integer coefficients, I would note that 5 factors only as 1(5) but neither 5-1 nor 5+ 1 is equal to 2 so we can't factor with integer coefficients. Not every quadratic can be factored with integer (or even real) coefficients.

So instead, complete the square! $x^2+ 2x+ 1= (x+ 1)^2$ so that $x^2+ 2x+ 5= x^2+ 2x+ 1+ 4= (x+1)^2+ 4= (x+1)^2- (-4)= (x+ 1)^2- (2i)^2$. Now use the identity $a^2- b^2= (a- b)(a+ b)$.
• Nov 1st 2012, 06:56 PM
Nervous
Re: Factoring help, again...
That's a beautiful idea... God, I still have so much to learn...
• Nov 2nd 2012, 05:20 AM
Nervous
Re: Factoring help, again...
Another one:
$2x^3+11x^2-12x-36$

I tried grouping, I tried factoring out an x and using quadratic formula on $x^+11x-12$ I don't know how to do this one.
• Nov 3rd 2012, 06:57 AM
Nervous
Re: Factoring help, again...
Edit
Quote:

Originally Posted by nervous
another one:
$2x^3+11x^2-12x-36$

i tried grouping, i tried factoring out an x and using quadratic formula on $x^2+11x-12$ i don't know how to do this one.

• Nov 3rd 2012, 03:16 PM
topsquark
Re: Factoring help, again...
Have you tried the "Rational Root Test?"

-Dan