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: 20

Answer

$\dfrac{3x}{4(3x+5)}; x \ne 0, -\frac{5}{3}$

Work Step by Step

Factor each polynomial completely to obtain: $=\dfrac{3}{2(x)} \cdot \dfrac{x(x)}{2(3x+5)}$ Cancel the common factors to obtain: $\require{cancel} \\=\dfrac{3}{2\cancel{(x)}} \cdot \dfrac{x\cancel{(x)}}{2(3x+5)} \\=\dfrac{3x}{4(3x+5)}; x \ne 0, -\frac{5}{3}$ ($x$ cannot be $0$ and $-\dfrac{5}{3}$ because they make the denominator equal to zero, which in turn makes the rational expression/s undefined.)
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