Finding the maclaurin series of a rational function confusion So I am told to find the maclaurin series of this function:
$f(x)=\frac{10}{x^2-2x-24}$ . The question is, how would I do that?
A few people told me to use partial fraction, geometric series or binomial series. I can see the reasoning, but I don't see why though. A Maclaurin series by definition is:

So I would imagine that I would just take my $f(x)$, put it inside the summation and be done with it right there. The question ask me to find any number of terms or anything...
It just says "Find the Maclaurin series for the following functions:"

$$f(x)=\frac{10}{(x-6)(x+4)} = \frac{1}{x-6}-\frac{1}{x+4}\tag{1}$$
and since in a neighbourhood of the origin
$$ \frac{1}{x-6}=-\frac{1}{6-x}=-\frac{1/6}{1-x/6}=-\sum_{n\geq 0}\frac{x^n}{6^{n+1}}, $$
$$ \frac{1}{4+x}=\frac{1/4}{1+x/4}=\sum_{n\geq 0}\frac{(-1)^n x^n}{4^{n+1}}\tag{2} $$
we have:
$$ f(x) = -\sum_{n\geq 0}\left(\frac{1}{6^{n+1}}+\frac{(-1)^n}{4^{n+1}}\right) x^n.\tag{3} $$
That is a power series with a positive convergence radius, hence it is the Taylor series of $f(x)$ at the origin (by the unicity of the Taylor series). In particular:
$$ f^{(n)}(0) = -n!\left(\frac{1}{6^{n+1}}+\frac{(-1)^n}{4^{n+1}}\right) \tag{4}$$
that is not that easy to get from repeated differentiation, unless $(1)$ is performed before.
That gives an alternative viable way: $(1)\mapsto RD\mapsto(4)\mapsto(3)$.
