pre algebraic factoring with polynomials I really need help solving this particular problem. 
$$\frac14x^2y(x-1)^3-\frac54xy(x-1)^2$$
I need help factoring this. It seems like I need to get rid of the fraction but I really just need a little boost.

Let's see, we have 
$$
\frac{1}{4}x^2y(x-1)^3-\frac{5}{4}xy(x-1)^2
$$
Notice the first term has three $x-1$'s and the second term has two of them. So they share two of them, so we can pull out two of them leaving
$$
(x-1)^2\left(\frac{1}{4}x^2y(x-1)^1-\frac{5}{4}xy\right)
$$
Notice, we 'used' two $x-1$ in the first term, leaving only one left and we took out all of them in the second. Anything else? Well, notice both terms have an $x$ and a $y$ in common. So let's take out one of each! This gives us
$$
xy(x-1)^2\left(\frac{1}{4}x^1(x-1)^1-\frac{5}{4}\right)
$$
The first term had only one $y$, which we took out and two $x$'s, taking one out leaves only one left. The second term had only one $x$ and one $y$ so we took both those out leaving nothing there. Are we done? Not quite. Notice the first term and the second term still have a $\frac{1}{4}$ in common, factoring that out yields
$$
\frac{1}{4}xy(x-1)^2\left(x^1(x-1)^1-5\right)
$$
Now all we have to do is clean this up a bit and we get the solution
$$
\frac{1}{4}xy(x-1)^2\left(x(x-1)-5\right)
$$
Some might even distribute the $x$ in $x(x-1)$ to get
$$
\frac{1}{4}xy(x-1)^2\left(x^2-x-5\right)
$$
as their answer because they would say it looks 'nicer'. 
