College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6 - Page 86: 31


$\frac{(x+4)(x+2)}{(x-5)}; x \ne -6,-3,-1,3,5$

Work Step by Step

Given expression is $\frac{x^{2}+x-12}{x^{2}+x-30} \times \frac{x^{2}+5x+6}{x^{2}-2x-3} \div \frac{x+3}{x^{2}+7x+6}$ Factor all the expressions in numerator and in denominator if possible. $=\frac{(x+4)(x-3)}{(x+6)(x-5)} \times \frac{(x+3)(x+2)}{(x-3)(x+1)} \div \frac{(x+3)}{(x+6)(x+1)} ; x \ne -6,-3,-1,3,5$ $ x \ne -6,-3,-1,3,5$ for non-zero denominators. $=\frac{(x+4)(x-3)}{(x+6)(x-5)} \times \frac{(x+3)(x+2)}{(x-3)(x+1)}\times\frac{(x+6)(x+1)}{(x+3)} ; x \ne -6,-3,-1,3,5$ Divide out the common factors. The result is $=\frac{(x+4)(x+2)}{(x-5)}; x \ne -6,-3,-1,3,5$
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