How to properly set up partial fractions for repeated denominator factors I was just trying to solve a problem that had the following item which I needed to split into separate generating functions:
$$\frac{x}{(1-2x)^2(1-5x)}$$
I had assumed I needed to split it into:
$$\frac{A}{1-2x} + \frac{B}{1-2x} + \frac{C}{1-5x}$$
But according to Wolfram Alpha it appears I had to split it into:
$$\frac{A}{1-2x} + \frac{B}{(1-2x)^2} + \frac{C}{1-5x}$$
Can anyone explain the intuition behind this? Is this a general rule that when you have a repeated factor in the denominator, you split it into all powers of that factor?
 A: Consider the simplest of cases. $$\frac{\xi}{x^3}$$ Assuming $\xi$ is some polynomial, one could carry on long division to determine the quotient and the remainder. Thus one could get $$\frac{\xi}{x^3}=q(x) + \frac{r(x)}{x^3}$$ Now what can we say for certain about $r(x)$ ?  We can say that it is the remainder thus of smaller degree than $x^3$. The most general possible polynomial of degree 2 or smaller is $A+Bx+Cx^2$. Then $$\frac{r(x)}{x^3}=\frac{A+Bx+Cx^2}{x^3}$$ Further more, we can split the right side $$\frac{r(x)}{x^3}=\frac{A}{x^3}+\frac{Bx}{x^3}+\frac{Cx^2}{x^3}$$ which becomes 
$$\frac{r(x)}{x^3}=\frac{A}{x^3}+\frac{B}{x^2}+\frac{C}{x}$$ Thus the above would be a sensible what to try to split any proper fraction $\frac{r(x)}{x^3}$ Moreover, the similar idea would hold for $\frac{r(x)}{(x+a)^3}$,  meaning a sensible way to split it would be $$\frac{r(x)}{(x+a)^3}=\frac{A}{(x+a)^3} +\frac{B}{(x+a)^2}+\frac{C}{(x+a)} $$  hope that helps.. 
A: Yep - if there's a factor of $(x-a)^k$ in the denominator, you need to include $\frac{1}{x-a},\frac{1}{(x-a)^2},...,\frac{1}{(x-a)^k}$ terms in your expansion. 
Note that in your attempt, you could combine the first two terms to $\frac{A+B}{1-2x}$, making one of those constants redundant.
A: Note that if $A$ and $B$ are constants then
$$\frac{A}{1-2x}+\frac{B}{1-2x}=\frac{A+B}{1-2x},$$
so that really these two terms are only one.
