College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.7 - Rational Expressions - R.7 Assess Your Understanding - Page 71: 19


$\dfrac{3}{5x(x-2)}; x \ne 0,-2, 2$

Work Step by Step

Factor each polynomial completely to obtain: $=\dfrac{3(x+2)}{5(x)(x)} \cdot \dfrac{x}{(x-2)(x+2)}$ Cancel the common factors to obtain: $\require{cancel} \\=\dfrac{3\cancel{(x+2)}}{5\cancel{(x)}(x)} \cdot \dfrac{\cancel{x}}{(x-2)\cancel{(x+2)}} \\=\dfrac{3}{5(x)(x-2)} \\=\dfrac{3}{5x(x-2)}; x \ne 0,-2, 2$ ($x$ cannot be $0$, $-2$, and $2$ because they make the denominator equal to zero, which in turn makes the rational expression undefined.)
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