Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Section 6.1 - Rational Functions and Multiplying and Dividing Rational Expressions - Exercise Set: 89

Answer

$b=\dfrac{(x+2)(x-1)^2}{x^5}$

Work Step by Step

Using $A=bh$ or the formula for the area of a parallelogram, then \begin{array}{l}\require{cancel} A=bh \\ \dfrac{x^2+x-2}{x^3}=b\left( \dfrac{x^2}{x-1} \right) \\ \dfrac{x^2+x-2}{x^3}\div\dfrac{x^2}{x-1}=b \\ b=\dfrac{x^2+x-2}{x^3}\cdot\dfrac{x-1}{x^2} \\ b=\dfrac{(x+2)(x-1)}{x^3}\cdot\dfrac{x-1}{x^2} \\ b=\dfrac{(x+2)(x-1)^2}{x^5} .\end{array}
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