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: 68


$\displaystyle \frac{-2(2x+7)}{(x-1)^{2}(x+2)^{2}}$

Work Step by Step

Step 1: Factor each denominator ... both are given in factored form Step 2: The LCM is the product of each of these factors raised to a power equal to the greatest number of times that the factor occurs in the polynomials. LCM = $(x-1)^{2}(x+2)^{2}$ Step 3: Write each rational expression using the LCM as the denominator. Simplify. $\displaystyle \frac{2}{(x+2)^{2}(x-1)}\cdot\frac{(x-1)}{(x-1)}-\frac{6}{(x+2)(x-1)^{2}}\cdot\frac{(x+2)}{(x+2)}$ $=\displaystyle \frac{2x-2-6x-12}{(x-1)^{2}(x+2)^{2}}$ $=\displaystyle \frac{-4x-14}{(x-1)^{2}(x+2)^{2}}$ $=\displaystyle \frac{-2(2x+7)}{(x-1)^{2}(x+2)^{2}}$
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