Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Appendix A - Review - A.6 Rational Expressions - A.6 Assess Your Understanding - Page A53: 24



Work Step by Step

By factoring both the denominator and numerator of each rational function and cancelling the common factors, we have $$ \frac{x^2+x-6}{x^2+4x-5}\frac{x^2-25}{x^2+2x-15}=\frac{(x+3)(x-2)}{(x+5)(x-1)}\frac{(x-5)(x+5)}{(x-3)(x+5)}\\ =\frac{(x+3)(x-2)(x-5)}{(x-3)(x-1)(x+5)} .$$
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