Binomial Expansion Of $\frac{24}{(x-4)(x+3)}$ Can somebody help me expand $\frac{24}{(x-4)(x+3)}$ by splitting it in partial fractions first and then using the general binomial theorem?
This is what I've done so far:
$$\frac{24}{(x-4)(x+3)}$$
$$=\frac{24}{7(x-4)}-\frac{24}{7(x+3)}$$
Now I know I have to find the binomial expansion for this; I just don't know how.
Can anyone help me with this?

$\dfrac{24}{7(x-4)}=\dfrac{-6}{7\left(1-\frac x4\right)}=-\dfrac67\left(1+\frac x4+(\frac x4)^2+(\frac x4)^3+\cdots\right)$
$\dfrac{24}{7(x-(-3))}=\dfrac{8}{7\left(1-\left(-\frac x3\right)\right)}=\dfrac87\left(1-\frac x3+(\frac x3)^2-(\frac x3)^3+\cdots\right)$
$\therefore\dfrac{24}{(x-4)(x+3)}=-2+\dfrac16x-\dfrac{13}{72}x^2 +\dfrac{25}{864}x^3\cdots$
