Homework question: Represent function as power series Represent the following function as a power series of powers of x-2.
$$f(x)=\frac{x-5}{3x^2+5x-2}$$
Now, the Taylor series got me thinking. Knowing my professor, she wouldn't have us solving huge derivatives like this one. So I'm wondering, is there any other way to solve this?
 A: Yes, there is another way of approaching this.
First, it will make things a bit easier to let $y=x-2$.  
Then, write 
$$\begin{align}
3x^2+5x-2&=(3x-1)(x+2)\\\\
&=(3(x-2)+5)((x-2)+4)\\\\
&=(3y+5)(y+4)
\end{align}$$  
Next, using partial fraction expansion reveals that
$$\begin{align}
\frac{x-5}{3x^2+5x-2}&=\frac{y-3}{(3y+5)(y+4)}\\\\
&=\frac{1}{y+4}+\frac{-2}{3y+5} \tag 1
\end{align}$$
Now, we can expand each term of the right-hand side of $(1)$ separately.  For the first term, we have 
$$\begin{align}
\frac{1}{y+4}&=\frac{1/4}{1+(y/4)}\\\\
&=\frac14\sum_{n=0}^{\infty}(-y/4)^n \tag 2
\end{align}$$
which is valid for $|y|<4$.
For the second term, we have
$$\begin{align}
\frac{-2}{3y+5}&=-\frac{2/5}{1+(3y/5)}\\\\
&=-\frac25 \sum_{n=0}^{\infty}(-3y/5)^n \tag 3
\end{align}$$
which is valid for $|y|<5/3$.
Now, adding $(2)$ and $(3)$ yields
$$\begin{align}
\frac{y-3}{(3y+5)(y+4)}=\sum_{n=0}^{\infty}\left(\frac{1}{4^{n+1}}-\frac{2(3)^n}{5^{n+1}}\right)(-1)^ny^n
\end{align}$$
whereupon substituting back $y=x-2$ yields
$$\begin{align}
\frac{x-5}{3x^2+5x-2}=\sum_{n=0}^{\infty}(-1)^n\left(\left(\frac{1}{4}\right)^{n+1}-\frac23\left(\frac{3}{5}\right)^{n+1}\right)(x-2)^n
\end{align}$$
where the series converges for $|x-2|<5/3$.

NOTE:
We can obtain a series representation for $|x-2|>5/3$ in terms of inverse powers of $x-2$.   That exercise is left to the reader.
A: Yes, you could factor the denominator and see what gives
