Finding a MacLaurin expansion of a function I am being asked to Find the MacLaurin expansion of the following function:
$f(x) = \frac{2x-8}{x^2-8x+12}$
I was not given a point about which to expand so I assume to use x=0. I know I can begin taking derivatives but this seems inefficient, but I do not see any way to reduce it to relate it to a more common series. Is there something I am missing, or is taking derivatives the only option?

$\begin{array}\\
f(x) 
&= \frac{2x-8}{x^2-8x+12}\\
&= \frac{2x-8}{(x-6)(x-2)}\\
&=2(x-4)\frac14\left( \frac1{x-6}-\frac1{x-2}\right)\\
&=\frac12(x-4)\left( \frac1{x-6}-\frac1{x-2}\right)\\
&=\frac12\left( \frac{x-4}{x-6}-\frac{x-4}{x-2}\right)\\
&=\frac12\left( \frac{x-6+2}{x-6}-\frac{x-2-2}{x-2}\right)\\
&=\frac12\left( 1+\frac{2}{x-6}-(1-\frac{2}{x-2})\right)\\
&=\frac12\left( \frac{2}{x-6}+\frac{2}{x-2}\right)\\
&= \frac{1}{x-6}+\frac{1}{x-2}\\
\end{array}
$
Now,
get the MacLaurin expansions
of
$\frac1{x-a}$
for
$a=2$ and $a=6$
and add them.
