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: 23

Answer

$\dfrac{x-3}{x+7} $

Work Step by Step

By factoring both the denominator and numerator of each rational function and cancelling the common factors, we have $$ \frac{x^2-3x-10}{x^2+2x-35}\frac{x^2+4x-21}{x^2+9x+14}=\frac{(x-5)(x+2)}{(x-5)(x+7)}\frac{(x-3)(x+7)}{(x+2)(x+7)}\\ =\frac{x-3}{x+7} .$$
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