Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.3 Rational Expressions - R.3 Exercises: 21

Answer

$\dfrac{k^{2}+4k-12}{k^{2}+10k+24}\cdot\dfrac{k^{2}+k-12}{k^{2}-9}=\dfrac{k-2}{k+3}$

Work Step by Step

$\dfrac{k^{2}+4k-12}{k^{2}+10k+24}\cdot\dfrac{k^{2}+k-12}{k^{2}-9}$ Factor both rational expressions completely: $\dfrac{k^{2}+4k-12}{k^{2}+10k+24}\cdot\dfrac{k^{2}+k-12}{k^{2}-9}=...$ $...=\dfrac{(k+6)(k-2)}{(k+6)(k+4)}\cdot\dfrac{(k+4)(k-3)}{(k-3)(k+3)}=...$ Evaluate the product of the two fractions: $...=\dfrac{(k+6)(k-2)(k+4)(k-3)}{(k+6)(k+4)(k-3)(k+3)}=...$ Simplify the resulting expression by removing the factors that appear both in the numerator and in the denominator: $...=\dfrac{k-2}{k+3}$
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