Answer
$\frac{2}{x+2}, x\ne4, x\ne-2$
Work Step by Step
Factoring $x^{2}-2x-8$,
$x^{2}-2x-8$
$=(x-4)(x+2)$
Factoring $2x-8$,
$2x-8$
$=2(x-4)$
$x-4$ can be cancelled on the top and bottom of the fraction, leaving
$\frac{2}{x+2}$
This assumes an important condition that $x\ne4$ (since you cannot cancel zero), and $x\ne-2$ (since $\frac{2}{0}$ is undefined).
