$$(3x-1)(2x+2)(8x+0) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -1 & x_3 = \dfrac{ 1 }{ 3 } \end{matrix} $$
Explanation
$$ \begin{aligned} (3x-1)(2x+2)(8x+0) &= 0&& \text{simplify left side} \\[1 em](6x^2+6x-2x-2)(8x+0) &= 0&& \\[1 em](6x^2+4x-2)(8x+0) &= 0&& \\[1 em]48x^3+0x^2+32x^2+0x-16x+0 &= 0&& \\[1 em]48x^3+0x^2+32x^2+0x-16x+0 &= 0&& \\[1 em]48x^3+32x^2-16x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 48x^{3}+32x^{2}-16x = 0 } $, first we need to factor our $ x $.
$$ 48x^{3}+32x^{2}-16x = x \left( 48x^{2}+32x-16 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 48x^{2}+32x-16 = 0$.
$ 48x^{2}+32x-16 = 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.
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