Answer
$\dfrac{2x(x^2+4x+16)}{x+4}, x\ne-4, 0, 4$
Work Step by Step
Not e that:
$\dfrac{4x^2}{x^2-16}\cdot \dfrac{x^3-64}{2x}=\dfrac{4x^2}{x^2-4^2}\cdot \dfrac{x^3-4^3}{2x}$
Factor each polynomial completely.
Use special formulas $a^3-b^3=(a-b)(a^2+ab+b^2)$ and $a^2-b^2=(a+b)(a-b)$ and determine the restrictions to obtain:
$=\dfrac{4x^2}{(x+4)(x-4)}\cdot \dfrac{(x-4)(x^2+4x+16)}{2x}, x\ne-4, 0, 4$
Cancel the common factors $x-4, 2, $ and $x$:
$=\dfrac{2x(x^2+4x+16)}{x+4}, x\ne-4, 0, 4$
Hence, the lowest term is:
$\dfrac{2x(x^2+4x+16)}{x+4}, x\ne-4, 0, 4$.
