Chapter 9 - Quadratic Functions and Equations - 9-4 Factoring to Solve Quadratic Equations - Practice and Problem-Solving Exercises - Page 572: 39

The equation has three solutions, $x=0$ or $x=4$ or $x=6$.

We are given $x^3-10x^2+24x=0$ $x^3-6x^2-4x^2+24x=0$ $(x^3-6x^2)-(4x^2-24x)=0$ $x^2(x-6)-4x(x-6)=0$ $(x^2-4x)(x-6)=0$ $x(x-4)(x-6)=0$ $x=0$ $x-4=0$ or $x-6=0$ $x=4$ or $x=6$ Therefore, the equation has three solutions, $x=0$ or $x=4$ or $x=6$