Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 7 - Section 7.2 - Multiplying and Dividing Rational Expressions - Exercise Set - Page 499: 49


$\displaystyle \frac{2(x+3)}{3}$

Work Step by Step

Dividing with $\displaystyle \frac{P}{Q}$ equals multiplying with the reciprocal, $\displaystyle \frac{Q}{P}.$ $ \displaystyle \frac{4x^{2}+10}{x-3}\div\frac{6x^{2}+15}{x^{2}-9}=\frac{4x^{2}+10}{x-3}\cdot\frac{x^{2}-9}{6x^{2}+15}\qquad$... factor what you can $\left[\begin{array}{l} 4x^{2}+10=2(x^{2}+5)\\ 6x^{2}+15=3(x^{2}+5)\\ x^{2}-9=(x+3)(x-3) \end{array}\right]$ $=\displaystyle \frac{2(x^{2}+5)}{(x-3)}\cdot\frac{(x+3)(x-3)}{3(x^{2}+5)}\qquad$... divide out the common factors $=\displaystyle \frac{2(1)}{(1)}\cdot\frac{(x+3)(1)}{3(1)}{\ }$ = $\displaystyle \frac{2(x+3)}{3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.