Answer
$-1, 2,3$
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$
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13446-914-3
ISBN 13:
978-0-13446-914-0
Chapter 4 - Cumulative Review Exercises - Page 647: 5
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