$$4x(2x-5)(x+4) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -4 & x_3 = \dfrac{ 5 }{ 2 } \end{matrix} $$
Explanation
$$ \begin{aligned} 4x(2x-5)(x+4) &= 0&& \text{simplify left side} \\[1 em](8x^2-20x)(x+4) &= 0&& \\[1 em]8x^3+32x^2-20x^2-80x &= 0&& \\[1 em]8x^3+12x^2-80x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 8x^{3}+12x^{2}-80x = 0 } $, first we need to factor our $ x $.
$$ 8x^{3}+12x^{2}-80x = x \left( 8x^{2}+12x-80 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 8x^{2}+12x-80 = 0$.
$ 8x^{2}+12x-80 = 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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