Chapter 11.2, Problem 28PPE

To determine

Multiply: $(w^2-8w+15)·\frac{w+3}{4w-20}$

Answer to Problem 28PPE

  $(w^2-8w+15)·\frac{w+3}{4w-20}=\frac{1}{4}(w^2-9)$

Explanation of Solution

Given:

Multiply: $(w^2-8w+15)·\frac{w+3}{4w-20}$

Concept Used:

For multiplication of rational expression, we always cancel the common factor from numerator and denominator and if necessary multiply the numerators and denominators separately.

Calculation:

  $\begin{array}{l}(w^2-8w+15)·\frac{w+3}{4w-20}\\ (w-5)(w-3)·\frac{w+3}{4(w-5)}\\ \frac{(w-3)(w+3)}{4}·\frac{(w-5)}{(w-5)}\\ \frac{1}{4}(w^2-9)\end{array}$

Thus, $(w^2-8w+15)·\frac{w+3}{4w-20}=\frac{1}{4}(w^2-9)$