Question

Resolve the following fraction into partial fraction: $\frac{x^2+7x}{x^2+7x-8}$.

Answer 1

$E=\frac{x^2+7x}{x^2+7x-8}$

Now,

$E=\frac{x^2+7x-8+8}{x^2+7x-8}=1+\frac{8}{x^2+7x-8}$.

$E$ can be rewritten using partial fractions.

$E=1+\frac{8}{x^2+7x-8}=1+(\frac{A}{x+8}+\frac{B}{x-1})$

Comparing both sides, we get

$A+B=0$ and $8B-A=8$

Therefore, $B=\frac{8}{9},A=-\frac{8}{9}$

$E=1+\frac{8}{9}(\frac{1}{x-1}-\frac{1}{x+8})$