$$2(x+1)^3(x-1)^2(x-2) = 0$$
Answer
Explanation
$ \color{blue}{ 2x^{6}-2x^{5}-8x^{4}+4x^{3}+10x^{2}-2x-4 } $ is a polynomial of degree 6. 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 (-4) are 1 2 4 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } ~ $$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} $
$$ \frac{ 2x^{6}-2x^{5}-8x^{4}+4x^{3}+10x^{2}-2x-4 }{ \color{blue}{ x - 1 } } = 2x^{5}-8x^{3}-4x^{2}+6x+4 $$Polynomial $ 2x^{5}-8x^{3}-4x^{2}+6x+4 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ 2x^{5}-8x^{3}-4x^{2}+6x+4 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.