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?
 A: $\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.
A: Taking partial fractions helps put it in a neater form. Here's my solution below:
\begin{align*}
\frac{2x-8}{x^2-8x+12}&=2\left(\frac{x-4}{(x-6)(x-2)}\right).
\end{align*}
Now we assume we can break this apart using partial fractions:
\begin{align*}
LHS&=2\left(\frac{a}{x-6}+\frac{b}{x-2}\right)\\
\end{align*}
Rewriting this as a single fraction gives
\begin{align*}
LHS&=2\left(\frac{a(x-2)+b(x-6)}{(x-6)(x-2)}\right)\\
&=2\left(\frac{(a+b)x-2a-6b)}{(x-6)(x-2)}\right)\\
\end{align*}
Equating coefficients, we see $a+b=1$, and $-2a-6b=-4$. Solving this system of questions give $a=b=\frac{1}{2}$. Substituting this back in for $a$ and $b$ gives
\begin{align*}
LHS&=2\left(\frac{1}{2(x-6)}+\frac{1}{2(x-2)}\right)\\
&=\frac{1}{x-6}+\frac{1}{x-2}
\end{align*}
Next we take the $n^\text{th}$ derivative of a function of the form $f=\frac{1}{x-a}$. We're going to use the mathematical principle of induction here to find a formula for all natural numbers. If you aren't familiar with this concept, we show something holds when $n=1$, then show that if it holds for some arbitrary value $k$ then $k+1$ also holds. 
For $n=1$, $f'=-\frac{1}{(x-a)^2}$. Assume that $f^{(k)}=\frac{(-1)^kk!}{(x-a)^{k+1}}$. Consider $f^{(k+1)}$. This is the same as
\begin{align*}
f^{(k+1)}&=\frac{d}{dx}(f^{(k)})
\end{align*}
From our induction hypothesis,
\begin{align*}
f^{(k+1)}&=\frac{d}{dx}\left(\frac{(-1)^kk!}{(x-a)^{k+1}}\right)\\
&=(-1)^kk!\frac{d}{dx}\left(\frac{1}{(x-a)^{k+1}}\right)\\
&=(-1)^kk!\left(\frac{-(k+1)(x-a)^{k}}{(x-a)^{2(k+1)}}\right)\\
&=\frac{(-1)^{k+1}(k+1)!}{(x-a)^{2(k+1)-k}}\\
&=\frac{(-1)^{k+1}(k+1)!}{(x-a)^{k+2}}
\end{align*}
Thus $k+1$ holds and $f^{(n)}=\frac{(-1)^nn!}{(x-a)^{n+1}}$ is true for all positive integers. Therefore, we can write the Maclaurian expansions of these two functions as
\begin{align*}
\frac{1}{x-6}+\frac{1}{x-2}&=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n+\sum_{n=0}^\infty\frac{g^{(n)}(0)}{n!}x^n\\
&=\sum_{n=0}^\infty\frac{f^{(n)}(0)+g^{(n)}(0)}{n!}x^n\\
&=\sum_{n=0}^\infty\frac{\frac{(-1)^nn!}{(-6)^{n+1}}+\frac{(-1)^nn!}{(-2)^{n+1}}}{n!}x^n\\
&=\sum_{n=0}^\infty\left(\frac{(-1)^n}{(-6)^{n+1}}+\frac{(-1)^n}{(-2)^{n+1}}\right)x^n\\
&=\sum_{n=0}^\infty\left(\frac{1}{6^{n+1}}+\frac{1}{2^{n+1}}\right)x^n
\end{align*}
