Question

Find each product or quotient.

$\frac{3b-12}{b+4}÷\left(b^2-6b+8\right)$

Short answer

The simplified expression is $\frac{3}{b^2+2b-8}$.

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 b2−6b+8.

First multiply by the reciprocal of $b^2-6b+8$and then factor the expressions in numerator and denominator.

role="math" localid="1648050245885" $$\begin{gathered} \frac{3b-12}{b+4}÷\left(b^2-6b+8\right)=\frac{3b-12}{b+4}\times \frac{1}{b^2-6b+8} \\ =\frac{3\left(b-4\right)}{b+4}\times \frac{1}{\left(b-4\right)\left(b-2\right)} \end{gathered}$$

Step 3. Simplify the expression.

Further simplify the expression by cancelling out the common factors $b-4$.

role="math" localid="1648050393663" $\begin{array}{c}\frac{3b-12}{b+4}÷\left(b^2-6b+8\right)=\frac{3\left(\cancel{b-4}\right)}{b+4}\times \frac{1}{\left(\cancel{b-4}\right)\left(b-2\right)}\\ =\frac{3}{\left(b+4\right)\left(b-2\right)}\\ =\frac{3}{b^2+2b-8}\end{array}$

Therefore, the simplified expression is role="math" localid="1648050403311" $\frac{3}{b^2+2b-8}$.