$$4x \cdot \frac{6x^3-8x^2+12-8}{12}x^2 = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -0.58891 & x_3 = 0.96112+0.45637i \\[1 em] x_4 = 0.96112-0.45637i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 4x \cdot \frac{6x^3-8x^2+12-8}{12}x^2 &= 0&& \text{multiply ALL terms by } \color{blue}{ 12 }. \\[1 em]124x \cdot \frac{6x^3-8x^2+12-8}{12}x^2 &= 12\cdot0&& \text{cancel out the denominators} \\[1 em]24x^4-32x^3+16x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 24x^{4}-32x^{3}+16x = 0 } $, first we need to factor our $ x $.
$$ 24x^{4}-32x^{3}+16x = x \left( 24x^{3}-32x^{2}+16 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 24x^{3}-32x^{2}+16 = 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
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