Chapter 9 - Quadratic Functions and Equations - 9-4 Factoring to Solve Quadratic Equations - Practice and Problem-Solving Exercises - Page 572: 44

The equation has three solutions, $x=-1$ or $x=2$ or $x=-2$

We are given $x^3+x^2-4x-4=0$ $(x^3+x^2)-(4x+4)=0$ $x^2(x+1)-4(x+1)=0$ $(x+1)(x^2-4)=0$ $(x+1)(x-2)(x+2)=0$ $x+1=0$ or $x-2=0$ or $x+2=0$ $x=-1$ or $x=2$ or $x=-2$ Therefore, the equation has three solutions, $x=-1$ or $x=2$ or $x=-2$