Chapter 6.6, Problem 4MRE

To determine

Simplify: the given expression as fraction.

Answer to Problem 4MRE

The simplified form of $\frac{n^2-4}{2}÷\frac{n+2}{8}$ is $4n-8$.

Explanation of Solution

Given information:

The expression $\frac{n^2-4}{2}÷\frac{n+2}{8}$.

Formula used:

To divide two fraction, the dividend fraction and the reciprocal of divisor fraction is multiplied.

  $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$

  $a^2-b^2=\left(a+b\right)\left(a-b\right)$.

  $\frac{n^2-4}{2}÷\frac{n+2}{8}$

Recall that to divide two fraction, the dividend fraction and the reciprocal of divisor fraction is multiplied.

  $=\frac{n^2-4}{2}\times \frac{8}{n+2}$

Now, factorizing numerator of dividend by using identity: $a^2-b^2=\left(a+b\right)\left(a-b\right)$.

  $\begin{array}{l}=\frac{n^2-2^2}{2}\times \frac{8}{n+2}\\ =\frac{\left(n-2\right)\left(n+2\right)}{2}\times \frac{8}{n+2}\end{array}$

Simplifying the expression by cancelling out the common factors.

  $\begin{array}{l}=4\left(n-2\right)\\ =4n-8\end{array}$

The simplified form of $\frac{n^2-4}{2}÷\frac{n+2}{8}$ is $4n-8$.