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?

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.
