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 5 - Polynomials and Factoring - Review Exercises: Chapter 5 - Page 370: 58



Work Step by Step

Solving the equation by factoring, we obtain: $$ 0=\left(x-2\right)\cdot \:2x^2+x\left(x-2\right)-\left(x-2\right)\cdot \:15 \\ \left(x-2\right)\left(x+3\right)\left(2x-5\right)=0 \\ x=2,\:x=-3,\:x=\frac{5}{2}$$
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