Factoring $4x^4 + 12x^3 - 24x^2 - 32x$ Some help with factorizing this polynomial please. I have tried but it is difficult as it factorizes down to a cubic and I can't factorize it further. This is regarding the division of polynomials.
$$P(x) = 4x^4 + 12x^3 - 24x^2 - 32x$$
 A: HINT:
Taking out $4x $ as common, $4x^4+12x^3-24x^2-32x=4x(x^3+3x^2-6x-8)$
Now, $3x^2-6x=3x(x-2), x^3-8=x^3-2^3$
A: Factor theorem will help.
$$4x^4 + 12x^3 - 24x^2 - 32x=4x(x^3+3x^2-6x-8).$$
Here, letting $f(x)=x^3+3x^2-6x-8,$ we have
$$f(-1)=0.$$
This implies that $f(x)$ has a factor $(x-(-1)).$
So, we have
$$4x(x+1)(x^2+2x-8).$$
Then, you'll know how to proceed from here.
A: You can factor $4 x$ first. Now, by inspection, you could find that $x=-1$ is a solution. Now, reduce from cubic to quadratic and continue.  
I am sure you can take from here.
A: Maybe this could help:
$$4x^4+12x^3-24x^2-32x=(4x^4-24x^2)+(12x^3-3x)=4x^2(x^2-6)+4x^2(-6-\frac8x)$$
Then we can sum the two terms
$$4x^2(x^2-12-\frac8x)=4x^2(\frac{x^3-12x-8}x)=4x(x^3-12x-8)=4x^4-48x-32x$$
Hope this would help
Peterix
A: Start by:
$$P(x) = 4x^4 + 12x^3 - 24x^2 - 32x=P(x) = 4x(x^3 + 3x^2 - 6x - 8)$$
Then one root is $x=-1$. So divide by $(x+1)$ and get simpler second order  polynomial. 
In general, if you interested in result rather than the way, much websites like Wolfram Alpha can solve this problems, doing for you the long and sometimes ugly way. All you need to do is to feed the polynomial and get this result saying it is $4 x (x+1) (x-2) (x+4)$.
