College Algebra (10th Edition)

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

Chapter 1 - Test: 6


$x=-2$ or $x=2$ or $x=-\displaystyle \frac{2}{3}$

Work Step by Step

We solve by factoring with grouping: $3x^{3}+2x^{2}-12x-8=0$ $x^{2}(3x+2)-4(3x+2)=0$ $(3x+2)(x^{2}-4)=0$ $(3x+2)(x+2)(x-2)=0$ $3x+2=0$ or $x+2=0$ or $x-2=0$ $x=-\displaystyle \frac{2}{3}$ or $x=-2$ or $x=2$
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