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: 65


$\displaystyle \frac{2(2x^{2}+5x-2)}{(x-2)(x+2)(x+3)}$

Work Step by Step

Step 1: Factor each denominator $ x^{2}-4=\qquad$ ... a difference of squares, $=(x-2)(x+2)$ For $x^{2}+bx+c$, we search for factors of c whose sum is b: $ x^{2}+x-6=\qquad$ ... +3 and -2 $=(x+3)(x-2)$ Step 2: The LCM is the product of each of these factors raised to a power equal to the greatest number of times that the factor occurs in the polynomials. LCM = $(x-2)(x+2)(x+3)$ Step 3: Write each rational expression using the LCM as the denominator. Simplify. $\displaystyle \frac{4x}{(x-2)(x+2)}\cdot\frac{(x+3)}{(x+3)}-\frac{2}{(x+3)(x-2)}\frac{(x+2)}{(x+2)}$ $=\displaystyle \frac{4x^{2}+12x-(2x+4)}{(x-2)(x+2)(x+3)}$ $=\displaystyle \frac{4x^{2}+10x-4}{(x-2)(x+2)(x+3)}$ $=\displaystyle \frac{2(2x^{2}+5x-2)}{(x-2)(x+2)(x+3)}$ For $ax^{2}+bx+c$ in the numerator, we search for factors of ac whose sum is b, and rewrite the $bx$ term: ... no two factors of -4 add to +5, so we leave it as it is.
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.