$$m(m^3-m^2+3) = 0$$
Answer
$$ \begin{matrix}m_1 = 0 & m_2 = -1.17456 & m_3 = 1.08728+1.17131i \\[1 em] m_4 = 1.08728-1.17131i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} m(m^3-m^2+3) &= 0&& \text{simplify left side} \\[1 em]m^4-m^3+3m &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{4}-x^{3}+3x = 0 } $, first we need to factor our $ x $.
$$ x^{4}-x^{3}+3x = x \left( x^{3}-x^{2}+3 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ x^{3}-x^{2}+3 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
This page was created using
Equations Solver