$$6 \cdot \frac{x^3}{2}x(2x+1) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 1 }{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 6 \cdot \frac{x^3}{2}x(2x+1) &= 0&& \text{simplify left side} \\[1 em]\frac{6x^3}{2}x(2x+1) &= 0&& \\[1 em]\frac{6x^4}{2}(2x+1) &= 0&& \\[1 em]\frac{12x^5+6x^4}{2} &= 0&& \text{multiply ALL terms by } \color{blue}{ 2 }. \\[1 em]2 \cdot \frac{12x^5+6x^4}{2} &= 2\cdot0&& \text{cancel out the denominators} \\[1 em]12x^5+6x^4 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 12x^{5}+6x^{4} = 0 } $, first we need to factor our $ x^4 $.
$$ 12x^{5}+6x^{4} = x^4 \left( 12x+6 \right) $$$ x = 0 $ is a root of multiplicity $ 4 $.
The second root can be found by solving equation $ 12x+6 = 0$.
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