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: 64


$\displaystyle \frac{x^{2}+7x-1}{(x-3)(x+8)}$

Work Step by Step

Step 1: Factor each denominator For $x^{2}+bx+c$, we search for factors of c whose sum is b: $ x^{2}+5x-24=\qquad$ ... +8 and -3 $=(x+8)(x-3)$ 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 = $(x+8)(x-3)$ Step 3: Write each rational expression using the LCM as the denominator. Simplify. $\displaystyle \frac{x(x+8)}{(x-3)(x+8)}-\frac{x+1}{(x+8)(x-3)}=$ $=\displaystyle \frac{x(x+8)-(x+1)}{(x-3)(x+8)}$ $=\displaystyle \frac{x^{2}+8x-x-1}{(x-3)(x+8)}$ $=\displaystyle \frac{x^{2}+7x-1}{(x-3)(x+8)}$ ... no two factors of -1 add to seven, so we leave the numerator as it is.
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