$$z(z+2)(z-1) = 0$$
Answer
$$ \begin{matrix}z_1 = 0 & z_2 = 1 & z_3 = -2 \end{matrix} $$
Explanation
$$ \begin{aligned} z(z+2)(z-1) &= 0&& \text{simplify left side} \\[1 em](1z^2+2z)(z-1) &= 0&& \\[1 em]z^3-z^2+2z^2-2z &= 0&& \\[1 em]z^3+z^2-2z &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{3}+x^{2}-2x = 0 } $, first we need to factor our $ x $.
$$ x^{3}+x^{2}-2x = x \left( x^{2}+x-2 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ x^{2}+x-2 = 0$.
$ x^{2}+x-2 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
This page was created using
Polynomial Equations Solver