College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6 - Page 86: 51


$\frac{4x+16}{(x+3)^{2}}; x \ne -3;$ or $\frac{4(x+4)}{(x+3)^{2}}; x \ne -3;$

Work Step by Step

$\frac{4}{x^{2}+6x+9} + \frac{4}{x+3}$ $x^{2}+6x+9$ is in the form of $A^{2}+2AB+B^{2} = (A+B)^{2}$ $x^{2}+6x+9 = (x+3)^{2}$ So, $=\frac{4}{ (x+3)^{2}} + \frac{4}{x+3}$ Taking Least Common Denominator, $=\frac{4+4(x+3)}{ (x+3)^{2}} ; x \ne -3 $ $=\frac{4+4x+12}{ (x+3)^{2}} ; x \ne -3 $ $= \frac{4x+16}{(x+3)^{2}}; x \ne -3;$ $=\frac{4(x+4)}{(x+3)^{2}}; x \ne -3;$
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.