Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 9 - Inequalities and Problem Solving - 9.4 Inequalities in Two Variables - 9.4 Exercise Set - Page 611: 63

Answer

$$\frac{\left(x+1\right)\left(3x+2\right)}{2\left(x+5\right)\left(2x-3\right)}$$

Work Step by Step

Recall, multiplying rational expressions means that you multiply the numerators by each other and also multiply the denominators by each other to get the new numerator and denominator, respectively. Thus, we find: $$ \frac{\left(15x^2-65x-50\right)\left(x^2+2x+1\right)}{\left(2x^2-x-3\right)\left(10x^2-250\right)}\\ \frac{\left(x+1\right)\left(x-5\right)\left(3x+2\right)}{2\left(x+5\right)\left(x-5\right)\left(2x-3\right)}\\\ \frac{\left(x+1\right)\left(3x+2\right)}{2\left(x+5\right)\left(2x-3\right)}$$
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