I'd like to bring again one of the below-mentioned proofs, but in a slightly modified form. In fact, I'll use the same words as you can read it in H. Wilfs Generatingfunctionology (section 1.2). The text is for your amusement and the benefit is that you could use these techniques, if you need an answer quickly.
We have the form
and the only problem is how to find the constants $A$, $B$, $C$.
Here's the quick way. First multiply both sides of $(\ast)$ by $(x+2)^2$ and then let $x=-2$. The instant result is that $C=-3$ (don't take my word for it, try it for yourself!). Next multiply $(\ast)$ through by $x-1$ and let $x=1$. The instant result is that $A=1$. The hard one to find is $B$, so let's do that one by cheating. Since we know that $(\ast)$ is an identity, i.e., is true for all values of $x$, let's choose an easy value of $x$, say $x=0$, and substitute that value of $x$ into $(\ast)$. Since we know $A$ and $C$, we find at once that $B=-1$.