# How do you find all the real and complex roots of f(x)= x^3-x+3?

May 24, 2016

Use Cardano's method to find Real zero:

${x}_{1} = \frac{1}{6} \left(\sqrt[3]{- 324 + 12 \sqrt{717}} + \sqrt[3]{- 324 - 12 \sqrt{717}}\right)$

and related Complex zeros.

#### Explanation:

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

Using the rational root theorem, we find that the only possible rational zeros of $f \left(x\right)$ are: $\pm 1$, $\pm 3$. None of these are zeros, so $f \left(x\right)$ has no rational zeros.

In fact it turns out to have one Real zero and two non-Real Complex zeros.

We can use Cardano's method to find all three as follows:

Let $x = u + v$. Then $f \left(x\right) = 0$ becomes:

${u}^{3} + {v}^{3} + \left(3 u v - 1\right) \left(u + v\right) + 3 = 0$

Let $v = \frac{1}{3 u}$ to eliminate the term in $\left(u + v\right)$ and get:

${u}^{3} + \frac{1}{27 {u}^{3}} + 3 = 0$

Multiply through by $27 {u}^{3}$ to get:

$27 {\left({u}^{3}\right)}^{2} + 81 \left({u}^{3}\right) + 1 = 0$

Using the quadratic formula, we get:

${u}^{3} = \frac{- 81 \pm \sqrt{{81}^{2} - 4 \cdot 27 \cdot 1}}{2 \cdot 27}$

$= \frac{- 81 \pm \sqrt{6561 - 108}}{54}$

$= \frac{- 81 \pm \sqrt{6453}}{54}$

$= \frac{- 81 \pm 3 \sqrt{717}}{54}$

$= \frac{- 324 \pm 12 \sqrt{717}}{216}$

Since these roots are Real and the derivation was symmetric in $u$ and $v$, we can use one of these roots for ${u}^{3}$ and the other for ${v}^{3}$ to find the Real zero:

${x}_{1} = \frac{1}{6} \left(\sqrt[3]{- 324 + 12 \sqrt{717}} + \sqrt[3]{- 324 - 12 \sqrt{717}}\right)$

and Complex zeros:

${x}_{2} = \frac{1}{6} \left(\omega \sqrt[3]{- 324 + 12 \sqrt{717}} + {\omega}^{2} \sqrt[3]{- 324 - 12 \sqrt{717}}\right)$

${x}_{3} = \frac{1}{6} \left({\omega}^{2} \sqrt[3]{- 324 + 12 \sqrt{717}} + \omega \sqrt[3]{- 324 - 12 \sqrt{717}}\right)$