Question

Resolve the following fraction into partial fractions : $\frac{16}{(x+2)(x^2-4)}$.

Answer 1

$\frac{16}{(x+2)(x^2-4)}=\frac{16}{(x+2)(x+2)(x-2)}$
$=\frac{16}{(x-2)(x+2{)}^2}$
Suppose that $\frac{16}{(x-2)(x+2{)}^2}=\frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{(x+2{)}^2}$
$\Rightarrow 16=A(x+2{)}^2+B(x-2\left)\right(x+2)+C(x-2).....(i)$
By Putting in equation (i)
$x=2$
$x=-2$
$A=1,C=-4$ again by putting, we get the value of $B=-1$.
$=\frac{16}{(x+2)(x^2-4)}=\frac{1}{x-2}-\frac{1}{x+2}-\frac{4}{(x+2{)}^2}$