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 $2$?

$$
2 \sum_{n=0}^{15} \left(-\frac x 2 \right)^n + 6x \sum_{n=0}^{15} (-x^2)^n
$$
The $15$th-degree term of this sum is $\dfrac{f^{(15)}(0)}{15!} x^{15}.$
It is also $2\left( -\dfrac x 2 \right)^{15} + 6x(-x^2)^7.$
From that one can find $f^{(15)}(0).$
