Chapter 11 - Quadratic Functions and Equations - 11.3 Studying Solutions of Quadratic Equations - 11.3 Exercise Set - Page 718: 56

$$0=x^3-3x^2-4x+12$$

To write a cubic equation with the given values as solutions, we find the corresponding factors. We then FOIL out the factors to find the original quadratic equation: $$ 0=\left(x+2\right)\left(x-2\right)\left(x-3\right) \\ 0=\left(x^2-4\right)\left(x-3\right) \\ 0=x^3-3x^2-4x+12$$