# Thread: How do I solve this polynomial equation?

1. ## How do I solve this polynomial equation?

The equation is 27x^4+x=0. If the equation was x^4+x^2=0 I could substitute u=x^2 and solve it as a regulator quadratic equation, but I don't know how to solve this one

2. ## Re: How do I solve this polynomial equation?

$27x^4 + x = 0$

$(27x^3 +1)x = 0$

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.

3. ## Re: How do I solve this polynomial equation?

Elegant, thanks a lot

4. ## Re: How do I solve this polynomial equation?

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.
Alternatively, as \displaystyle \begin{align*} 27\,x^3 + 1 &= \left( 3\,x \right) ^3 + 1^3 \end{align*}, a sum of two cubes, it factorises easily as \displaystyle \begin{align*} \left( 3\,x \right) ^3 + 1^3 &= \left( 3\,x + 1 \right) \left[ \left( 3\,x \right) ^2 - 3\,x \cdot 1 + 1^2 \right] \end{align*}...