Chapter 9 - Quadratic Functions and Equations - 9-4 Factoring to Solve Quadratic Equations - Practice and Problem-Solving Exercises - Page 572: 45

The equation has three solutions, $x=-2$ or $x=3$ or $x=-3$.

We are given $x^3+2x^2-9x-18=0$ $(x^3+2x^2)-(9x+18)=0$ $x^2(x+2)-9(x+2)=0$ $(x+2)(x^2-9)=0$ $(x+2)(x-3)(x+3)=0$ $x+2=0$ or $x-3=0$ or $x+3=0$ $x=-2$ or $x=3$ or $x=-3$ Therefore, the equation has three solutions, $x=-2$ or $x=3$ or $x=-3$