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

$$ \begin{matrix}x_1 = 1 & x_2 = -1 & x_3 = -3 \end{matrix} $$

$ \color{blue}{ 2x^{6}+16x^{5}+34x^{4}-16x^{3}-90x^{2}+54 } $ 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 (54) are 1 2 3 6 9 18 27 54 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 2 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 18 }{ 2 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 2 } , ~ \pm \frac{ 54 }{ 1 } , ~ \pm \frac{ 54 }{ 2 } ~ $$

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} $

$$ \frac{ 2x^{6}+16x^{5}+34x^{4}-16x^{3}-90x^{2}+54 }{ \color{blue}{ x - 1 } } = 2x^{5}+18x^{4}+52x^{3}+36x^{2}-54x-54 $$

Polynomial $ 2x^{5}+18x^{4}+52x^{3}+36x^{2}-54x-54 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ 2x^{5}+18x^{4}+52x^{3}+36x^{2}-54x-54 $

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

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