Chapter 5 - Section 5.8 - Solving Equations by Factoring and Problem Solving - Exercise Set - Page 326: 112

$4$ and $-3$ are solutions of the quadratic equation $x^2-x-12=0$.

RECALL: If $(x-a)(x-b)=0$, then $x=a$ and $x=b$ are solutions of the equation $(x-a)(x-b)=0$. Since $4$ and $-3$ are solutions of the quadratic equation, then the equation must be: $(x-4)(x+3)=0$ Multiply the binomials to have: $\\x(x+3) -4(x+3)=0 \\x^2+3x-4x-12=0 \\x^2-x-12=0$