$$5x^3(2x-8) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 4 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 5x^3(2x-8) &= 0&& \text{simplify left side} \\[1 em]10x^4-40x^3 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 10x^{4}-40x^{3} = 0 } $, first we need to factor our $ x^3 $.
$$ 10x^{4}-40x^{3} = x^3 \left( 10x-40 \right) $$$ x = 0 $ is a root of multiplicity $ 3 $.
The second root can be found by solving equation $ 10x-40 = 0$.
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