find $f^{(15)}(0)$ if $f = \frac{10x^2+12x+4}{(x+2) (x^2 +1)}$ $$f = \frac{10x^2+12x+4}{(x+2)(x^2 +1)}=\frac{4}{x+2}+\frac{6x}{x^2+1}=\frac{2}{1-\left(-\frac x 2 \right)}+\frac{6x}{1-(-x^2)}$$
So we end up with $$2 \sum_{n=0}^{15} \left(-\frac x 2 \right)^n + 6x \sum_{n=0}^{15} (-x^2)^n$$ but because we are looking for $f^{(15)}(0)$ we get the result of $2$?

If you have a taylor polynomial for $f(x)$ i.e. $f(x) = \sum a_n x^n$ then
$f^{(n)}(0) = n! a_n$
Normally we think about this the other way around.  We want to find the coefficients and we say $a_n = \frac {f^{(n)}(0)}{n!}$ 
$f(x) = 2\sum (-\frac 12)^n x^n + 6x\sum (-1)^n x^{2n}\\
a_{15} = 2 (-\frac 12)^{15} + 6 (-1)^7\\
 -2^{-14} - 6$
$f^{(15)}(0) = (-2^{-14} - 6)15!$
