$$2x^2(x-4)^3(x+1) = 0$$

$$ \begin{matrix}x_1 = 0 & x_2 = -1 & x_3 = 4 \end{matrix} $$

In order to solve $ \color{blue}{ 2x^{6}-22x^{5}+72x^{4}-32x^{3}-128x^{2} = 0 } $, first we need to factor our $ x^2 $.

$$ 2x^{6}-22x^{5}+72x^{4}-32x^{3}-128x^{2} = x^2 \left( 2x^{4}-22x^{3}+72x^{2}-32x-128 \right) $$

$ x = 0 $ is a root of multiplicity $ 2 $.

The remaining roots can be found by solving equation $ 2x^{4}-22x^{3}+72x^{2}-32x-128 = 0$.

$ \color{blue}{ 2x^{4}-22x^{3}+72x^{2}-32x-128 } $ is a polynomial of degree 4. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $ \dfrac{p}{q} $, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

The factors of the leading coefficient ( 2 ) are 1 2 .The factors of the constant term (-128) are 1 2 4 8 16 32 64 128 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 16 }{ 2 } , ~ \pm \frac{ 32 }{ 1 } , ~ \pm \frac{ 32 }{ 2 } , ~ \pm \frac{ 64 }{ 1 } , ~ \pm \frac{ 64 }{ 2 } , ~ \pm \frac{ 128 }{ 1 } , ~ \pm \frac{ 128 }{ 2 } ~ $$

Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial $ P(x) $, we obtain $ P(-1) = 0 $.

To find remaining zeros we use Factor Theorem. This theorem states that if $\frac{p}{q}$ is root of the polynomial then this polynomial can be divided with $ \color{blue}{q x - p} $. In this example:

Divide $ P(x) $ with $ \color{blue}{x + 1} $

$$ \frac{ 2x^{4}-22x^{3}+72x^{2}-32x-128 }{ \color{blue}{ x + 1 } } = 2x^{3}-24x^{2}+96x-128 $$

Polynomial $ 2x^{3}-24x^{2}+96x-128 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ 2x^{3}-24x^{2}+96x-128 $

When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.

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