# Appendix A - Review - A.6 Rational Expressions - A.6 Assess Your Understanding - Page A53: 19

$\dfrac{2x(x^2+4x+16)}{x+4}$

#### Work Step by Step

By factoring both the denominator and numerator of each rational function and cancelling the common factors, we have $$\frac{4x^2}{x^2-16}\cdot \frac{x^3-64}{2x} = \frac{4x^2}{(x-4)(x+4)}\cdot \frac{(x-4)(x^2+4x+16)}{2x} \\ = \frac{2x(x^2+4x+16)}{x+4} .$$

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