Chapter 9 - Quadratic Functions and Equations - 9-4 Factoring to Solve Quadratic Equations - Practice and Problem-Solving Exercises - Page 572: 40

The equation has three solutions, $x=0$ or $x=4$ or $x=1$

We are given $x^3-5x^2+4x=0$ $x^3-4x^2-x^2+4x=0$ $(x^3-4x^2)-(x^2-4x)=0$ $x^2(x-4)-x(x-4)=0$ $(x^2-x)(x-4)=0$ $x(x-4)(x-1)=0$ $x=0$ or $x-4=0$ or $x-1=0$ $x=4$ or $x=1$ Therefore, the equation has three solutions, $x=0$ or $x=4$ or $x=1$