Answer
$\frac{12}{(x-3)(x+3)}=\frac{2}{x-3}+\frac{-2}{x+3}$
Work Step by Step
$\frac{12}{(x^2-9)}$, factorising $x^2-9=(x-3)(x+3)$:
$\frac{12}{(x-3)(x+3)}=\frac{A}{x-3}+\frac{B}{x+3}$,
$12=A(x+3)+B(x-3)$,
$12=Ax+3A+Bx-3B$,
$12=(A+B)x+(3A-3B)$,
$(A+B)x=0x$,
$A+B=0$ and $3A-3B=12, A-B=4$:
$\begin{cases}
& A & -B & = 4&\\
& A & +B & = 0&\\
& -- & -- & -- \\
& 2A & +0 & = 4
\end{cases}$
$A=2$, substituting back into the equation, $2+B=0, B=-2$
thus,
$\frac{12}{(x-3)(x+3)}=\frac{2}{x-3}+\frac{-2}{x+3}$
