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: 22

Answer

$\dfrac{6(x^2-x+1)}{x(2x-1)};x \ne 0,-1,\frac{1}{2}$

Work Step by Step

Factor each polynomial completely to obtain: $=\dfrac{2(2)(3)}{x(x+1)} \cdot \dfrac{(x+1)(x^2-x+1)}{2(2x-1)}$ Cancel the common factors to obtain: $\require{cancel} \\=\dfrac{2\cancel{(2)}(3)}{x\cancel{(x+1)}} \cdot \dfrac{\cancel{(x+1)}(x^2-x+1)}{\cancel{2}(2x-1)} \\=\dfrac{2(3)(x^2-x+1)}{x(2x-1)} \\=\dfrac{6(x^2-x+1)}{x(2x-1)};x \ne 0,-1,\frac{1}{2}$ ($x$ cannot be $-1$, $0$, and $\dfrac{1}{2}$ because they make the denominator equal to zero, which in turn makes the rational expression/s undefined.)
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