## College Algebra (10th Edition)

Published by Pearson

# Chapter R - Section R.7 - Rational Expressions - R.7 Assess Your Understanding - Page 71: 70

#### Answer

$\displaystyle \frac{x^{2}-6+11}{(x+7)(x+1)^{2}}$

#### Work Step by Step

Step 1: Factor each denominator $x^{2}+8x+7=...$ For $x^{2}+bx+c$, we search for factors of c whose sum is b: ... $+7$ and $+1$ are such factors, $=(x+7)(x+1)$ $(x+1)^{2}=\qquad$... is already factored. 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 = $(x+7)(x+1)^{2}$ Step 3: Write each rational expression using the LCM as the denominator. Simplify. $\displaystyle \frac{2x-3}{(x+7)(x+1)}\cdot\frac{(x+1)}{(x+1)}-\frac{x-2}{(x+1)^{2}}\cdot\frac{(x+7)}{(x+7)}$ $= \displaystyle \frac{2x^{2}+2x-3x-3-(x^{2}+5x-14)}{(x+7)(x+1)^{2}}$ $=\displaystyle \frac{x^{2}-6+11}{(x+7)(x+1)^{2}}$ For $ax^{2}+bx+c$ in the numerator, we search for factors of ac whose sum is b, and rewrite the $bx$ term: ... no two factors of $+11$ add to $-6$, so we leave it as it is.

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