Can someone help me with factoring after I derive a function?

I need to get the point of inflection so, the second derivative.
f(x)=x(x-4)^3
I got this far:
f'(x)=x(3)(x-4)^2+(x-4)^3
=3x(x^2-8x+16)+(x^2-8x+16)(x-4)
=3x^3-24x^2+48x)+(x^3-12x^2+48x-64
=4x^3-36x^2+96x-64
=4(x^3-9x^2+24x-16)
This is where I get stuck...
I mean, if you know an easier way?..

Question

Can someone help me with factoring after I derive a function? 

I need to get the point of inflection so, the second derivative. 
f(x)=x(x-4)^3
I got this far: 
f'(x)=x(3)(x-4)^2+(x-4)^3
=3x(x^2-8x+16)+(x^2-8x+16)(x-4)
=3x^3-24x^2+48x)+(x^3-12x^2+48x-64
=4x^3-36x^2+96x-64
=4(x^3-9x^2+24x-16)
This is where I get stuck... 
I mean, if you know an easier way?..

zepdrix · Answer

Woops! After "I got this far:" the second step after that, bad bad bad. Don't expand out those terms :O messy messy! :O
Notice they both some amount of (x-4)'s?
Factor out the largest power that they each share. (Looks like it's (x-4)^2 in this case).

zepdrix · Answer

|dw:1351899845811:dw|
From here you have a couple of nice easy critical points you can find. Understand what happened there? :D