College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Section P.7 - Rational Expressions - P.7 Exercises: 33

Answer

$\frac{x+5}{(2x+3)(x+4)}$

Work Step by Step

We switch from division to multiplication by taking the reciprocal of the right fraction. Then we factor and simplify: $\displaystyle \frac{x+3}{4x^{2}-9}\div\frac{x^{2}+7x+12}{2x^{2}+7x-15}=\frac{x+3}{4x^{2}-9}*\frac{2x^{2}+7x-15}{x^{2}+7x+12}=\frac{x+3}{(2x-3)(2x+3)}*\frac{(x+5)(2x-3)}{(x+3)(x+4)}=\frac{x+5}{(2x+3)(x+4)}$
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