Answer
$\dfrac{3}{5x(x-2)}; x \ne 0,-2, 2$
Work Step by Step
Factor each polynomial completely to obtain:
$=\dfrac{3(x+2)}{5(x)(x)} \cdot \dfrac{x}{(x-2)(x+2)}$
Cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{3\cancel{(x+2)}}{5\cancel{(x)}(x)} \cdot \dfrac{\cancel{x}}{(x-2)\cancel{(x+2)}}
\\=\dfrac{3}{5(x)(x-2)}
\\=\dfrac{3}{5x(x-2)}; x \ne 0,-2, 2$
($x$ cannot be $0$, $-2$, and $2$ because they make the denominator equal to zero, which in turn makes the rational expression undefined.)
