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.
 A: The given equation is
$$\dfrac{1}{4}x^2y(x - 1)^3 - \dfrac{5}{4}xy(x - 1)^2$$
Since the GCF (greatest common factor) is $\frac{1}{4}xy(x - 1)^2$, write the expression as
$$\begin{aligned}
\underbrace{\dfrac{1}{4}xy(x - 1)^2 \cdot x(x - 1)}_{\frac{1}{4}x^2y(x - 1)^3} - \underbrace{\dfrac{1}{4}xy(x - 1)^2 \cdot 5}_{\frac{5}{4}xy(x - 1)^2}&= \underbrace{\dfrac{1}{4}xy(x - 1)^2}_{\text{GCF}}\left(x(x - 1) - 5 \right)\\
&= \dfrac{1}{4}xy(x - 1)^2(x^2 - x - 5)
\end{aligned}$$
So the answer is
$$\dfrac{1}{4}xy(x - 1)^2(x^2 - x - 5)$$
Explanations for each step I have shown for your problem


*

*Extract the common factor out from each term you see, which is $\frac{1}{4}xy(x - 1)^2$ (Noticed that both terms have common terms $\frac{1}{4}$, $x$, $y$ and $(x - 1)^2$).  This is called the GCF.  Underneath each factored term, you see the original terms in the original equation.

*Factor out that term, leaving the uncommon factors (namely $x(x - 1)$ and $-5$) in the parentheses. (Multiple each term in the parentheses by $\frac{1}{4}xy(x - 1)^2$.  You should get the same answer.)

*Simplify the expression in the parentheses.

A: 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'. 
A: 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,
$$\frac{1}{4}xy(x-1)^2 \cdot (x(x-1)-5)=\frac{1}{4}xy(x-1)^2(x^2-x-5)$$
