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

$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 1 }{ 2 } & x_3 = -3 \end{matrix} $$

In order to solve $ \color{blue}{ 4x^{6}+46x^{5}+192x^{4}+324x^{3}+108x^{2}-162x = 0 } $, first we need to factor our $ x $.

$$ 4x^{6}+46x^{5}+192x^{4}+324x^{3}+108x^{2}-162x = x \left( 4x^{5}+46x^{4}+192x^{3}+324x^{2}+108x-162 \right) $$

$ x = 0 $ is a root of multiplicity $ 1 $.

The remaining roots can be found by solving equation $ 4x^{5}+46x^{4}+192x^{3}+324x^{2}+108x-162 = 0$.

$ \color{blue}{ 4x^{5}+46x^{4}+192x^{3}+324x^{2}+108x-162 } $ is a polynomial of degree 5. 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 ( 4 ) are 1 2 4 .The factors of the constant term (-162) are 1 2 3 6 9 18 27 54 81 162 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 4 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 6 }{ 4 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 2 } , ~ \pm \frac{ 9 }{ 4 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 18 }{ 2 } , ~ \pm \frac{ 18 }{ 4 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 2 } , ~ \pm \frac{ 27 }{ 4 } , ~ \pm \frac{ 54 }{ 1 } , ~ \pm \frac{ 54 }{ 2 } , ~ \pm \frac{ 54 }{ 4 } , ~ \pm \frac{ 81 }{ 1 } , ~ \pm \frac{ 81 }{ 2 } , ~ \pm \frac{ 81 }{ 4 } , ~ \pm \frac{ 162 }{ 1 } , ~ \pm \frac{ 162 }{ 2 } , ~ \pm \frac{ 162 }{ 4 } ~ $$

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 }{ 2 }) = 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}{ 2 x - 1 } $

$$ \frac{ 4x^{5}+46x^{4}+192x^{3}+324x^{2}+108x-162 }{ \color{blue}{ 2x - 1 } } = 2x^{4}+24x^{3}+108x^{2}+216x+162 $$

Polynomial $ 2x^{4}+24x^{3}+108x^{2}+216x+162 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ 2x^{4}+24x^{3}+108x^{2}+216x+162 $

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

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