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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?

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

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

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

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  • $\begingroup$ Could you give me the final answer for the whole fraction please? $\endgroup$ – mikejacob Jul 16 '20 at 16:31
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    $\begingroup$ Okay, I edited to include more details, @mikejacob $\endgroup$ – J. W. Tanner Jul 16 '20 at 17:02
  • $\begingroup$ Thank you very much, this is what i initially got but my book says the answer -7/4 + 7x/48 - 91x²/576 $\endgroup$ – mikejacob Jul 16 '20 at 17:16
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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 $|y|<1$ $$(1-y)^{-1}=\sum_{r=0}^\infty y^r$$

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