College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 83

Answer

$x=3$ or $x=2$ or $x=-2$

Work Step by Step

We solve by factoring with grouping: $x^{3}-3x^{2}-4x+12=0$ $x^{2}(x-3)-4(x-3)=0$ $(x-3)(x^{2}-4)=0$ $(x-3)(x-2)(x+2)=0$ $x-3=0$ or $x-2=0$ or $x+2=0$ $x=3$ or $x=2$ or $x=-2$
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