Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 11 - Rational Expressions and Functions - Chapter Test - Page 707: 9


$\dfrac{n^2+4n-5}{n+5} =n-1; n \ne -5$

Work Step by Step

Given: $\dfrac{n^2+4n-5}{n+5}$ Consider denominator $n+5=0$ to check the exclude value which makes the denominator $0$. $n+5=0$ Thus, $n=-5$ $\dfrac{n^2+4n-5}{n+5} = \dfrac{(n+5)(n-1)}{(n+5)}$ Hence, $\dfrac{n^2+4n-5}{n+5} =n-1; n \ne -5$
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