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: 31


$\dfrac{(x+3)^2}{(x-3)^2} $

Work Step by Step

We apply Method 1 (A51), to treat the numerator and denominator of the complex rational expression separately. So by factoring both the denominator and numerator of each rational function and cancelling the common factors, we have $$ \frac{\frac{x^2+7x+12}{x^2-7x+12}}{\frac{x^2+x-12}{x^2-x-12}}=\frac{\frac{(x+3)(x+4)}{(x-3)(x-4)}}{\frac{(x-3)(x+4)}{(x+3)(x-4)}}=\frac{(x+3)(x+4)}{(x-3)(x-4)}\frac{(x+3)(x-4)}{(x-3)(x+4)}\\ =\frac{(x+3)^2}{(x-3)^2} .$$
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