Consider the function $ f(z)=\frac{z}{1-z-z^2}$, I'm trying to determine the power series for the function by using partial fraction decomposition. I tried to expand it into $$f(z)= \frac{A}{(z-\alpha)}+\frac{B}{(z-\beta)}=\frac{1}{\alpha-\beta}\left( \frac{\alpha}{(z-\alpha)}-\frac{\beta}{(z-\beta)} \right)=\frac{1}{\alpha-\beta} \left( \frac{1}{1-\frac{z}{\beta}}-\frac{1}{1-z / \alpha}\right) $$ And then use the geometric series from there. But this doesn't quite give the desired results (Or at least it's not obvious that it's equivalent).
Instead, I looked up other solutions, where they expanded $f(z)=\frac{A}{1-z\alpha} + \frac{B}{1-z\beta}$
But I have never seen this kind of expansion before. Where does this come from? why is $1-z-z^2=(1-z\alpha)(1-z\beta)$? And can we factor all polynomials in such way (Because that is completely new to me)?
$\alpha,\beta$ are roots of the denominator of $f(z)$.