Question

Find each product or quotient.

$\frac{x^2}{x+4}÷\frac{3x}{x^2-16}$

Short answer

The simplified expression is $\frac{x^2-4x}{3}$.

Step-by-step solution

Step 1. Define the concept.

Multiplying rational expression: - Let a, b, c, and d be polynomials with$b\ne 0$ and $d\ne 0$. Then $\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$.

Dividing rational expression: - Let a, b, c, and d be polynomials with $b\ne 0$,$c\ne 0$ and $d\ne 0$. Then $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\cdot \frac{d}{c}=\frac{ad}{bc}$.

Step 2. Multiply by the reciprocal of 3xx2−16.

First multiply by the reciprocal of $\frac{3x}{x^2-16}$and then factor the expressions in numerator and denominator.

role="math" localid="1648049221815" $$\begin{gathered} \frac{x^2}{x+4}÷\frac{3x}{x^2-16}=\frac{x^2}{x+4}\times \frac{x^2-16}{3x} \\ =\frac{x^2}{x+4}\times \frac{\left(x+4\right)\left(x-4\right)}{3x} \end{gathered}$$

Step 3. Simplify the expression.

Further simplify the expression by cancelling out the common factors $x+4$ and x.

role="math" localid="1648049342460" $\begin{array}{c}\frac{x^2}{x+4}÷\frac{3x}{x^2-16}=\frac{x\times \cancel{x}}{\cancel{x+4}}\times \frac{\cancel{\left(x+4\right)}\left(x-4\right)}{3\cancel{x}}\\ =\frac{x\left(x-4\right)}{3}\\ =\frac{x^2-4x}{3}\end{array}$

Therefore, the simplified expression is $\frac{x^2-4x}{3}$.