# What are the zero(s) for f(x) = 2x^6 +x^3+3?

Oct 24, 2015

$f \left(x\right)$ has six Complex zeros which we can find by recognising that $f \left(x\right)$ is a quadratic in ${x}^{3}$.

#### Explanation:

$f \left(x\right) = 2 {x}^{6} + {x}^{3} + 3 = 2 {\left({x}^{3}\right)}^{2} + {x}^{3} + 3$

Using the quadratic formula we find:

${x}^{3} = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \times 2 \times 3}}{2 \cdot 2}$

$= \frac{- 1 \pm \sqrt{- 23}}{4} = \frac{- 1 \pm i \sqrt{23}}{4}$

So $f \left(x\right)$ has zeros:

${x}_{1 , 2} = \sqrt[3]{\frac{- 1 \pm i \sqrt{23}}{4}}$

${x}_{3 , 4} = \omega \sqrt[3]{\frac{- 1 \pm i \sqrt{23}}{4}}$

${x}_{5 , 6} = {\omega}^{2} \sqrt[3]{\frac{- 1 \pm i \sqrt{23}}{4}}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of unity.