College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.7 - Rational Expressions - R.7 Assess Your Understanding - Page 71: 33


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

Work Step by Step

$=\displaystyle \frac{x^{2}+7x+12}{x^{2}-7x+12}\div\frac{x^{2}+x-12}{x^{2}-x-12}=\frac{x^{2}+7x+12}{x^{2}-7x+12}\cdot\frac{x^{2}-x-12}{x^{2}+x-12}$ In each trinomial $x^{2}+bx+c$, we search for factors of c whose sum is b: $x^{2}+7x+12$ ... $,\quad $(4 and 3)$\quad $factors to $(x+4)(x+3)$ $x^{2}-7x+12$ ... $,\quad $($-4$ and $-3$)$\quad $factors to $(x-4)(x-3)$ $x^{2}-x-12$... $,\quad $($-4$ and $+3$)$\quad $factors to $(x-4)(x+3)$ $x^{2}+x-12$... $,\quad $($+4$ and $-3$)$\quad $factors to $(x+4)(x-3)$ $=\displaystyle \frac{(x+4)(x+3)(x-4)(x+3)}{(x-4)(x-3)(x+4)(x-3)}$ after canceling $(x+4)$ and $(x-4),$ $=\displaystyle \frac{(x+3)^{2}}{(x-3)^{2}}$
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.