Originally Posted by
romsek $27x^4 + x = 0$
$(27x^3 +1)x = 0$
so $x=0$ is one solution
now let's look at
$27x^3+1=0$
$x^3 = -\dfrac{1}{27}$
$x = -\dfrac 1 3$
is another solution
and dividing $27x^3+1$ by $\left(x + \dfrac 1 3\right)$ we get
$\dfrac{27x^3+1}{x + \frac 1 3} =27 x^2-9 x+3 $
now you can apply the quadratic formula to obtain the roots of $27x^2 -9x + 3$
$r_1,r_2 = \dfrac{9\pm \sqrt{81-4(27)(3)}}{2(27)} = \dfrac{9 \pm i 9 \sqrt{3}}{54} = \dfrac 1 6\left(1 \pm i \sqrt{3}\right)$
$0, -\dfrac 1 3, r_1, r_2$ are the 4 solutions to this quartic polynomial equation.