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I know how to factor a quadratic expression of the form $ax^2+bx+c$ into the form: $a(x - \frac{-b-\sqrt{b^2-4ac}}{2a})(x - \frac{-b+\sqrt{b^2-4ac}}{2a})$ using the root formulas. However, currently I'm learning about generating functions and it seems that it is advantageous for the factorization to be of the form $(1-ax)(1-bx)$, as $\frac{1}{1+kx}$ is the generating function for the sequence $a_n = k^x$. Can any quadratic expression be factored this way? How to do it?

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    $\begingroup$ No because this would require the constant term to always be 1 (expand it out). You could however always get the form $k(1-ax)(1-bx)$ $\endgroup$ – Osama Ghani Aug 16 '17 at 11:58
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You have say $$f(x)=\frac{g(x)}{(x-r)(x-s)}.$$ Then $$f(x)=\frac{r^{-1}s^{-1}g(x)}{(r^{-1}x-1)(s^{-1}x-1)} =\frac{r^{-1}s^{-1}g(x)}{(1-ux)(1-vx)}$$ where $u=1/r$, $v=1/s$.

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