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.

Maybe this will help you to see things clearer, but it is the same concept as @NasuSama.. 
$$\underbrace{\frac14x^2y(x-1)^3}_\text{one expression}-\underbrace{\frac54xy(x-1)^2}_\text{another expression}$$
Let $z=(x-1)$, your expression now becomes:
$$\frac{1}{4}x^2yz^3 - \frac{5}{4}xyz^2$$
You can easily see now that $\frac{1}{4}, x, y, \text{and}, z^2$, are common in both expressions..and so you can "factor out" $\frac{1}{4}xyz^2$ from both expressions. What you are doing when you are "factoring out" is trying to find what expression when multiplied by the term you are "factoring out" $(\frac{1}{4}xyz^2)$, will give you the original expression $(\frac{1}{4}x^2yz^3 - \frac{5}{4}xyz^2)$. This is equivalent to dividing $\frac{1}{4}x^2yz^3 - \frac{5}{4}xyz^2$ by $\frac{1}{4}xyz^2$. So dividing:
$$\frac{{\frac14x^2yz^3}-{\frac54xyz^2}}{\frac{1}{4}xyz^2}=\frac{\frac14x^2yz^3}{\frac{1}{4}xyz^2}-\frac{\frac54xyz^2}{\frac{1}{4}xyz^2}=xz-5$$
Hence
$$\underbrace{\frac{1}{4}xyz^2}_{a} \cdot {(xz-5)} =\underbrace{\frac{1}{4}x^2yz^3 - \frac{5}{4}xyz^2}_{b}$$
Once you have figured out what is common in both expressions, $a$, you simply need to figure out what you can multiply $a$ by to get $b$. Once you get more accustomed to factoring this will come more naturally. 
Now re substitute $z=(x-1)$ and you get,
$$\frac{1}{4}xy(x-1)^2 \cdot (x(x-1)-5)=\frac{1}{4}xy(x-1)^2(x^2-x-5)$$
