Chapter 9, Problem 15MS

To determine

To Simplify: The expression and assume that no denominator is zero.

Answer to Problem 15MS

  $\frac{(r-6){(r+6)}^2}{(r-3){(r+3)}^2}$

Explanation of Solution

Given: $\frac{r^2-r-20}{r^2-6r+5}\cdot \frac{r^2-36}{r^2-9}÷\frac{r^2+7r+12}{r^2+5r-6}$

Given: $\frac{r^2-r-20}{r^2-6r+5}\cdot \frac{r^2-36}{r^2-9}÷\frac{r^2+7r+12}{r^2+5r-6}$

Rewrite as multiplication and solve

  $\begin{array}{l}=\frac{r^2-r-20}{r^2-6r+5}\cdot \frac{r^2-36}{r^2-9}÷\frac{r^2+7r+12}{r^2+5r-6}\\ =\frac{r^2-r-20}{r^2-6r+5}\cdot \frac{r^2-36}{r^2-9}\cdot \frac{r^2+5r-6}{r^2+7r+12}\\ =\frac{(r-5)(r+4)}{(r-5)(r-1)}\times \frac{(r+6)(r-6)}{(r+3)(r-3)}\times \frac{(r-1)(r+6)}{(r+4)(r+3)}\\ =\frac{(r-6){(r+6)}^2}{(r-3){(r+3)}^2}\end{array}$

Hence, $\frac{(r-6){(r+6)}^2}{(r-3){(r+3)}^2}$