Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises - Page 138: 22

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

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 $x=-3$