Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises - Page 43: 27

Answer

$\frac{x-3}{x+2}$

Work Step by Step

$\frac{x^2+2x-15}{x^2-25}\times\frac{x-5}{x+2}$ Simplify $\frac{x^2+2x-15}{x^2-25}$: $=\frac{(x+5)(x-3)}{(x-5)(x+5)}$ $=\frac{(x-3)}{(x-5)}$ So it becomes: $=\frac{x-3}{x-5}\times\frac{x-5}{x+2}$ Multiply the two fractions (note: $x-5$ cancels out.): $=\frac{x-3}{x+2}$
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