<s>[INST] Find a power series expansion of $\frac{4x^2+2x}{1-3x-10x^2}$ about the point $x = \frac{1}{5}$ Find a power series expansion of $\frac{4x^2+2x}{1-3x-10x^2}$ 
Now I know that $\frac{1}{5}$ is a singularity of the $\frac{4x^2+2x}{1-3x-10x^2}$ 
and I know that $f(z) = f\left(\frac{1}{5}\right)+\frac{f'\left(\frac{1}{5}\right)}{1!}\left(x-\frac{1}{5}\right) + \ldots$ 
Now my question is this, how do I go about calculating $f\left(\frac{1}{5}\right)$, $f'\left(\frac{1}{5}\right)$, and so on.  
 [/INST] Factor your expression and you have 
$$\begin{align}
\frac{2x(2x+1)}{(1-5x)(1+2x)}
&=\frac{2x}{1-5x}\\
&=-{\frac{1}{5}}\frac{2\left(x-\frac15\right)+\frac25}{x-\frac15}\\
&=-{\frac{2}{5}}-\frac2{25}\left(x-\frac15\right)^{-1}
\end{align}$$
Note that computing a true power series about $\frac15$ is impossible precisely because there is a singularity there. So this is a Laurent series with just two terms.
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