Chapter 9 - Quadratic Functions and Equations - 9-4 Factoring to Solve Quadratic Equations - Practice and Problem-Solving Exercises - Page 572: 43

The equation has three solutions, $x=-5$ or $x=1$ or $x=-1$.

We are given $x^3+5x^2-x-5=0$ $(x^3+5x^2)-(x+5)=0$ $x^2(x+5)-(x+5)=0$ $(x+5)(x^2-1)=0$ $(x+5)(x-1)(x+1)=0$ $x+5=0$ or $x-1=0$ or $x+1=0$ $x=-5$ or $x=1$ or $x=-1$ Therefore, the equation has three solutions, $x=-5$ or $x=1$ or $x=-1$.