Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 7 - Review: 13

Answer

$\dfrac{x^{2}+xa+xb+ab}{x^{2}-xc+bx-bc}=\dfrac{x+a}{x-c}$

Work Step by Step

$\dfrac{x^{2}+xa+xb+ab}{x^{2}-xc+bx-bc}$ Group the first two terms and the last two terms in the numerator together. Do the same in the denominator: $\dfrac{x^{2}+xa+xb+ab}{x^{2}-xc+bx-bc}=\dfrac{(x^{2}+xa)+(xb+ab)}{(x^{2}-xc)+(bx-bc)}=...$ Take out common factor $x$ from the first parentheses in both the numerator and the denominator and common factor $b$ from the second parentheses in both the numerator and the denominator: $...=\dfrac{x(x+a)+b(x+a)}{x(x-c)+b(x-c)}=...$ Take out common factor $x+a$ from the numerator and common factor $x-c$ from the denominator and simplify: $...=\dfrac{(x+a)(x+b)}{(x+b)(x-c)}=\dfrac{x+a}{x-c}$
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