Answer
$\frac{x-5}{x+4}$ and $x\ne-1$.
Work Step by Step
$\frac{x^2-4x-5}{x^2+5x+4}=\frac{x^2+x-5x-5}{x^2+x+4x+4}=\frac{x(x+1)-5(x+1)}{x(x+1)+4(x+1)}=\frac{(x+1)(x-5)}{(x+1)(x+4)}=\frac{x-5}{x+4}$ and $x\ne-1$.
Intermediate Algebra for College Students (7th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13417-894-7
ISBN 13:
978-0-13417-894-3
Chapter 6 - Section 6.1 - Rational Expressions and Functions; Multiplying and Dividing - Exercise Set - Page 414: 40
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