Answer
$\frac{x-12}{x(x-4)}=\frac{3}{x}+\frac{-2}{x-4}$
Work Step by Step
$\frac{x-12}{(x^2-4x)}$, factorising $x^2-4x=x(x-4)$:
$\frac{x-12}{x(x-4)}=\frac{A}{x}+\frac{B}{x-4}$,
$x-12=A(x-4)+B(x)$,
$x-12=Ax-4A+Bx$,
$x-12=(A+B)x-4A$,
$(A+B)x=x$,
$A+B=1$ and $-4A=-12, A=3$:
$A=3$, substituting back into the equation, $3+B=1, B=-2$
thus,
$\frac{x-12}{x(x-4)}=\frac{3}{x}+\frac{-2}{x-4}$
