$$x^4(x-2)(x-16) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 2 & x_3 = 16 \end{matrix} $$
Explanation
$$ \begin{aligned} x^4(x-2)(x-16) &= 0&& \text{simplify left side} \\[1 em](x^5-2x^4)(x-16) &= 0&& \\[1 em]x^6-16x^5-2x^5+32x^4 &= 0&& \\[1 em]x^6-18x^5+32x^4 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{6}-18x^{5}+32x^{4} = 0 } $, first we need to factor our $ x^4 $.
$$ x^{6}-18x^{5}+32x^{4} = x^4 \left( x^{2}-18x+32 \right) $$$ x = 0 $ is a root of multiplicity $ 4 $.
The remaining roots can be found by solving equation $ x^{2}-18x+32 = 0$.
$ x^{2}-18x+32 = 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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