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: 32


$\frac{7(x+1)^{2}}{2x^{2}}; x \ne -5,-1,0,1,5$

Work Step by Step

The given expression can be factored as $=\frac{x(x^{2}-25)}{4x^{2}} \times \frac{2(x^{2}-1)}{(x-5)(x-1)} \div \frac{x(x+5)}{7(x+1)} ; x \ne -5,-1,0,1,5$ $ (x^{2}-25)$ and $(x^{2}-1) $ are difference of squares, so we can further factor as $= \frac{x(x+5)(x-5)}{4x^{2}} \times \frac{2(x-1)(x+1)}{(x-5)(x-1)} \div \frac{x(x+5)}{7(x+1)} ; x \ne -5,-1,0,1,5$ $x \ne -5,-1,0,1,5$, otherwise they make the denominators zero. $= \frac{x(x+5)(x-5)}{4x^{2}} \times \frac{2(x-1)(x+1)}{(x-5)(x-1)} \times\frac{7(x+1)}{x(x+5)} ; x \ne -5,-1,0,1,5$ Divide out common factors, the result is $=\frac{7(x+1)^{2}}{2x^{2}}; x \ne -5,-1,0,1,5$
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