I know factoring questions are a dime a dozen but I can't seem to get this one.
$-2x^3+2x^2+32x+40$
Factor to obtain the following equation:
$-2(x-5)(x+2)^2$
Do I have to use division (I'd prefer not to)? The way the question is worded, it seems I should just be able to pull factors out. This is the farthest I could make it:
$-2(x^3-x^2-16x-20)$
$-2[x^2(x-1)-4(4x+5)]$
$-2[(x^2-4)(x-1)(4x+5)]$
On
You can find the roots by the rational root theorem. Once you have pulled out the common $2$, the rational roots are factors of $20$, of which there are only $12$ (counting plus and minus). So just try them until you find one that works.
Once we know a factor, we can divide without dividing.
Start from $-2x^3+2x^2+32x+40$, and use the factor $x+2$.
We can rewrite as $(-2x^3-4x^2)+6x^2+32x+40$, and then as $(-2x^3-4x^2)+(6x^2+12x)+20x+40$.
So our expression is equal to $-2x^2(x+2)+6x(x+2)+20(x+2)$ and we can pull out the $x+2$. We get $(x+2)(-2x^2+6x+20)$.
But there is no good reason to avoid polynomial division!
Remark: If you are calculus-minded, let $P(x)$ be our polynomial. Write the Taylor expansion of $P(x)$ about $a=-2$, that is, in powers of $x+2$. Instead of division, we will use differentiation.