$$27p^4+3p = 2p$$

$$ \begin{matrix}p_1 = 0 & p_2 = -\dfrac{ 1 }{ 3 } & p_3 = \dfrac{ 1 }{ 6 }+\dfrac{\sqrt{ 3 }}{ 6 }i \\[1 em] p_4 = \dfrac{ 1 }{ 6 }-\dfrac{\sqrt{ 3 }}{ 6 }i & \\[1 em] \end{matrix} $$

In order to solve $ \color{blue}{ 27x^{4}+x = 0 } $, first we need to factor our $ x $.

$$ 27x^{4}+x = x \left( 27x^{3}+1 \right) $$

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

The remaining roots can be found by solving equation $ 27x^{3}+1 = 0$.

$ \color{blue}{ 27x^{3}+1 } $ 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 ( 27 ) are 1 3 9 27 .The factor of the constant term ( 1 ) is 1 . Then the Rational Roots Tests yields the following possible solutions:

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

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 } $

$$ \frac{ 27x^{3}+1 }{ \color{blue}{ 3x + 1 } } = 9x^{2}-3x+1 $$

Polynomial $ 9x^{2}-3x+1 $ can be used to find the remaining roots.

$ \color{blue}{ 9x^{2}-3x+1 } $ 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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