Answer
$\frac{x+14}{(x-4)(x+2)}=\frac{3}{x-4}+\frac{-2}{x+2}$
Work Step by Step
$\frac{x+14}{(x^2-2x-8)}$, factorising $x^2-2x-8=(x-4)(x+2)$:
$\frac{x+14}{(x-4)(x+2)}=\frac{A}{x-4}+\frac{B}{x+2}$,
$x+14=A(x+2)+B(x-4)$,
$x+14=Ax+2A+Bx-4B$,
$x+14=(A+B)x+(2A-4B)$,
$(A+B)x=x$,
$A+B=1$ and $2A-4B=14, A-2B=7$:
multiplying by $-1$ the first equation, we can solve for $B$.
$\begin{cases}
& A & -2B & = 7&\\
& -A & -B & = -1&\\
& -- & -- & -- \\
& 0 & -3B & = 6
\end{cases}$
$B=-2$, substituting back into the equation, $A-2=1, A=3$
thus,
$\frac{x+14}{(x-4)(x+2)}=\frac{3}{x-4}+\frac{-2}{x+2}$
