Binomial Expansion Of $\frac{24}{(x-4)(x+3)}$

Question

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?

Answer 1

$\dfrac{24}{7(x-4)}=\dfrac{-6}{7\left(1-\frac x4\right)}=-\dfrac67\left(1+\frac x4+(\frac x4)^2+(\frac x4)^3+\cdots\right)$

$\dfrac{24}{7(x-(-3))}=\dfrac{8}{7\left(1-\left(-\frac x3\right)\right)}=\dfrac87\left(1-\frac x3+(\frac x3)^2-(\frac x3)^3+\cdots\right)$

$\therefore\dfrac{24}{(x-4)(x+3)}=-2+\dfrac16x-\dfrac{13}{72}x^2 +\dfrac{25}{864}x^3\cdots$

Answer 2

Case$\#1:$ For $\left|\dfrac xr\right|<1,$

$\dfrac1{x-r}=-\dfrac1{r\left(1-\dfrac xr\right)}=-\dfrac1r\left(1-\dfrac xr\right)^{-1}$

Case$\#2:$ For $\left|\dfrac xr\right|>1\iff \left|\dfrac rx\right|<1,$

$\dfrac1{x-r}=\dfrac1{x\left(1-\dfrac rx\right)}=\dfrac1x\cdot\left(1-\dfrac rx\right)^{-1}$

Finally Using Binomial series, for $|y|<1$ $$(1-y)^{-1}=\sum_{r=0}^\infty y^r$$