Answer
$\dfrac{2x(x^2+4x+16)}{x+4};x \ne -4, 0, 4$
Work Step by Step
Factor each polynomial completely to obtain:
$=\dfrac{2(2)(x)(x)}{(x-4)(x+4)} \cdot \dfrac{(x-4)(x^2+4x+16)}{2(x)}$
Cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{2\cancel{(2)}\cancel{(x)}(x)}{\cancel{(x-4)}(x+4)} \cdot \dfrac{\cancel{(x-4)}(x^2+4x+16)}{\cancel{2}\cancel{(x)}}
\\=\dfrac{2x(x^2+4x+16)}{x+4};x \ne -4, 0, 4$
($x$ cannot be $-4$, $0$, and $4$ because they make the denominator equal to zero, which in turn makes the rational expression/s undefined.)
