finding the coefficient of $x^{14}$ in the expression: $\frac{5x^2-x^4}{(1-x)^3}$ I have a homework question which requires me to find the coefficient of $x^{14}$ in the expression: $\dfrac{5x^2-x^4}{(1-x)^3}$
I have not figured out a way to do this (I believe this is because my algebra is weak).
This is a question in a combinatorics class and has to do with Generating Functions if that is any help.
Help is greatly appreciated. Thanks, Jason

To complement @Bill's answer, I will explain one way to quickly obtain a partial fraction expansion in this example. 
Use substitution. Set $x = 1 -y$, so that the given expression becomes:
$$
\frac{5(1-y)^2 - (1-y)^4}{y^3} = \frac{4 - 6y - y^2 + 4y^3 - y^4}{y^3} =
\frac{4}{y^3} - \frac{6}{y^2} - \frac{1}{y} + 4 - y.
$$
Substituting back $1-x$ for $y$, we get:
$$
\frac{4}{(1-x)^3} - \frac{6}{(1-x)^2} - \frac{1}{(1-x)} + 3 + x.
$$
Using the binomial series1, we can calculate the coefficient of $x^{14}$ to be:
$$
4 \binom{14+3-1}{14} - 6 \binom{14+2-1}{14} - \binom{14+1-1}{14} + 0 + 0 = 4 \binom{16}{2} - 6 \times 15 - 1 = 389.
$$
1As I remarked in a comment above, the binomial series for $(1-x)^{\beta+1}$ has a particularly simple form. 
