Answer
$=4(x-2),\qquad x\neq-2,2$
Work Step by Step
Factor what we can. Recognize a difference of squares.
$\displaystyle \frac{x^{2}-4}{x-2}\div\frac{x+2}{4x-8}=\frac{(x-2)(x+2)}{x-2}\div\frac{x+2}{4(x-2)}$
Division with $\displaystyle \frac{a}{b}$ equals multiplication with $\displaystyle \frac{b}{a}.$
(neither a or b can be 0, we exclude $-2$ and $2$)
$=\displaystyle \frac{(x-2)(x+2)}{x-2}\cdot\frac{4(x-2)}{x+2}$
...exclude any additional values that yield 0 in the denominator:
$x\neq-2,2$
... cancel common factors
$=\displaystyle \frac{4(x-2)}{1},\qquad x\neq-2,2$
$=4(x-2),\qquad x\neq-2,2$
