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

$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 3 } & x_2 = -\dfrac{ 1 }{ 3 } & x_3 = 2 \end{matrix} $$

$ \color{blue}{ 81x^{6}-270x^{5}+108x^{4}+210x^{3}+23x^{2}-20x-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 ( 81 ) are 1 3 9 27 81 .The factors of the constant term (-4) are 1 2 4 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 1 }{ 9 } , ~ \pm \frac{ 1 }{ 27 } , ~ \pm \frac{ 1 }{ 81 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 3 } , ~ \pm \frac{ 2 }{ 9 } , ~ \pm \frac{ 2 }{ 27 } , ~ \pm \frac{ 2 }{ 81 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 3 } , ~ \pm \frac{ 4 }{ 9 } , ~ \pm \frac{ 4 }{ 27 } , ~ \pm \frac{ 4 }{ 81 } ~ $$

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(\frac{ 1 }{ 3 }) = 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}{ 3 x - 1 } $

$$ \frac{ 81x^{6}-270x^{5}+108x^{4}+210x^{3}+23x^{2}-20x-4 }{ \color{blue}{ 3x - 1 } } = 27x^{5}-81x^{4}+9x^{3}+73x^{2}+32x+4 $$

Polynomial $ 27x^{5}-81x^{4}+9x^{3}+73x^{2}+32x+4 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ 27x^{5}-81x^{4}+9x^{3}+73x^{2}+32x+4 $

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

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