College Algebra (10th Edition)

Published by Pearson

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

Answer

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

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 $(x-4),$ $=\displaystyle \frac{(x+3)^{2}}{(x-3)^{2}}$

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