$$3x^4 = 81x$$

$$ \begin{matrix}x_1 = 0 & x_2 = 3 & x_3 = -\dfrac{ 3 }{ 2 }+\dfrac{ 3 \sqrt{ 3}}{ 2 }i \\[1 em] x_4 = -\dfrac{ 3 }{ 2 }-\dfrac{ 3 \sqrt{ 3}}{ 2 }i & \\[1 em] \end{matrix} $$

In order to solve $ \color{blue}{ 3x^{4}-81x = 0 } $, first we need to factor our $ x $.

$$ 3x^{4}-81x = x \left( 3x^{3}-81 \right) $$

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

The remaining roots can be found by solving equation $ 3x^{3}-81 = 0$.

$ \color{blue}{ 3x^{3}-81 } $ is a polynomial of degree 3. 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 ( 3 ) are 1 3 .The factors of the constant term (-81) are 1 3 9 27 81 . Then the Rational Roots Tests yields the following possible solutions:

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

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(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}{x - 3} $

$$ \frac{ 3x^{3}-81 }{ \color{blue}{ x - 3 } } = 3x^{2}+9x+27 $$

Polynomial $ 3x^{2}+9x+27 $ can be used to find the remaining roots.

$ \color{blue}{ 3x^{2}+9x+27 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

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