Chapter 11.3, Problem 64MR

To determine

To multiply:

  $\frac{n^2+7n-8}{n-1}\cdot \frac{n^2-4}{n^2+6n-16}$

Answer to Problem 64MR

  $\frac{n^2+7n-8}{n-1}\cdot \frac{n^2-4}{n^2+6n-16}=n+2$

Explanation of Solution

Given:

  $\frac{n^2+7n-8}{n-1}\cdot \frac{n^2-4}{n^2+6n-16}$

Concept Used:

To multiply/divide rational expressions first simplify numerator and denominator of each rational expression and then proceed to find the operation by canceling the common terms of numerator and denominator.

Calculation:

In order to multiply:

  $\frac{n^2+7n-8}{n-1}\cdot \frac{n^2-4}{n^2+6n-16}$

First, simplify the numerator and denominator of each rational expression of given multiplication, as

  $\begin{array}{c}\frac{n^2+7n-8}{n-1}\cdot \frac{n^2-4}{n^2+6n-16}=\frac{n^2+8n-n-8}{n-1}\cdot \frac{n^2-2^2}{n^2+8n-2n-16}\\ =\frac{n\left(n+8\right)-\left(n+8\right)}{n-1}\cdot \frac{\left(n-2\right)\left(n+2\right)}{n\left(n+8\right)-2\left(n+8\right)}\\ =\frac{\left(n+8\right)\left(n-1\right)}{n-1}\cdot \frac{\left(n-2\right)\left(n+2\right)}{\left(n+8\right)\left(n-2\right)}\end{array}$

Now, cancel out the common terms of numerator and denominator, as

  $\frac{\left(n+8\right)\left(n-1\right)}{n-1}\cdot \frac{\left(n-2\right)\left(n+2\right)}{\left(n+8\right)\left(n-2\right)}=\frac{\left(n+8\right)}{1}\cdot \frac{\left(n+2\right)}{\left(n+8\right)}$

Multiplying numerator and denominator of rational expressions, as

  $\begin{array}{c}\frac{\left(n+8\right)}{1}\cdot \frac{\left(n+2\right)}{\left(n+8\right)}=\frac{\left(n+8\right)\left(n+2\right)}{\left(n+8\right)}\\ =\frac{\left(n+2\right)}{1},\left(\begin{array}{l}\text{cancelling the }\\ \text{common factor}\end{array}\right)\\ =n+2\end{array}$

Thus,

  $\frac{n^2+7n-8}{n-1}\cdot \frac{n^2-4}{n^2+6n-16}=n+2$