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: 5


$-1, 2,3$

Work Step by Step

Re-arrange as: $ x^3 -4x^2+x+6=0$ $\implies (x^3-2x^2)-(2x^2-4x)-(3x-6)=0$ $\implies x^2(x-2)-2x(x-2)-3(x-2)=0$ or, $(x-2)[x(x+1)-3 (x+1)]=0$ or, $(x-2) (x+1) (x-3)=0$ So, $ x=-1, 2,3$
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