$$(4m^3+2m)(8m^3-5m+4) = 0$$
Answer
$$ \begin{matrix}m_1 = 0 & m_2 = -1.04949 & m_3 = 0.52474+0.44841i \\[1 em] m_4 = 0.52474-0.44841i & m_5 = 0.70711i & m_6 = -0.70711i \end{matrix} $$
Explanation
$$ \begin{aligned} (4m^3+2m)(8m^3-5m+4) &= 0&& \text{simplify left side} \\[1 em]32m^6-20m^4+16m^3+16m^4-10m^2+8m &= 0&& \\[1 em]32m^6-4m^4+16m^3-10m^2+8m &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 32x^{6}-4x^{4}+16x^{3}-10x^{2}+8x = 0 } $, first we need to factor our $ x $.
$$ 32x^{6}-4x^{4}+16x^{3}-10x^{2}+8x = x \left( 32x^{5}-4x^{3}+16x^{2}-10x+8 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 32x^{5}-4x^{3}+16x^{2}-10x+8 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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