College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises: 22

Answer

$x=-\frac{1}{2}$ or $x=3$ or $x=-3$

Work Step by Step

We solve with factoring by grouping: $2x^{3}+x^{2}-18x-9=0$ $x^{2}(2x+1)-9(2x+1)=0$ $(2x+1)(x^{2}-9)=0$ $(2x+1)(x-3)(x+3)=0$ $2x+1=0$ or $x-3=0$ or $x+3=0$ $x=-\frac{1}{2}$ or $x=3$ or $x=-3$
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