Answer
$4(x-2); x \ne 2,-2$
Work Step by Step
$\frac{(x^{2}-4)}{x-2} \div \frac{x+2}{4x-8}$
Factorize $(x^{2}-4)$ as $(x+2)(x-2)$ and $(4x-8)$ as $4(x-2).$
$=\frac{(x+2)(x-2)}{x-2} \div \frac{x+2}{4(x-2)}; x \ne 2$
$=\frac{(x+2)(x-2)}{x-2} \times \frac{4(x-2)}{x+2}; x \ne 2,-2$
$=\frac{4(x+2)(x-2)(x-2)}{(x-2)(x+2)} ; x \ne 2,-2$
Divide out the common factors.
$=4(x-2); x \ne 2,-2$
