Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Cumulative Review Exercises - Page 647: 2



Work Step by Step

Re-arrange as: $ x^3+5x^2-4x-20=0$ $\implies x^2(x+5)-4(x+5)=0$ $\implies (x+5)(x^2-4)=0$ or, $(x+5)(x-2)(x+2)=0$ So, $ x=-5,-2,2$
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