# What are the zeros of x^3-8x-4?

Aug 11, 2016

${x}_{k} = \frac{4}{3} \sqrt{6} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{3}{16} \sqrt{6}\right) + \frac{2 k \pi}{3}\right) \text{ } k = 0 , 1 , 2$

#### Explanation:

$f \left(x\right) = {x}^{3} - 8 x - 4$

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Descriminant

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our example, $a = 1$, $b = 0$, $c = - 8$ and $d = - 4$, so we find:

$\Delta = 0 + 2048 + 0 - 432 + 0 = 1616$

Since $\Delta > 0$ this cubic has $3$ Real zeros.

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Trigonometric method

We use a substitution $x = k \cos \theta$, where $k$ is chosen to squeeze $f \left(x\right)$ into a form containing $4 {\cos}^{3} \theta - 3 \cos \theta = \cos 3 \theta$.

Let $k = \frac{4}{3} \sqrt{6}$

Then:

$0 = f \left(x\right) = {x}^{3} - 8 x - 4$

$= {\left(k \cos \theta\right)}^{3} - 8 \left(k \cos \theta\right) - 4$

$= k \left({k}^{2} {\cos}^{3} \theta - 8 \cos \theta\right) - 4$

$= \frac{4}{3} \sqrt{6} \left(\frac{32}{3} {\cos}^{3} \theta - 8 \cos \theta\right) - 4$

$= \frac{32}{9} \sqrt{6} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) - 4$

$= \frac{32}{9} \sqrt{6} \cos 3 \theta - 4$

Add $4$ to both ends and transpose to get:

$\frac{32}{9} \sqrt{6} \cos 3 \theta = 4$

Multiply both sides by $\sqrt{6}$ to get:

$\frac{64}{3} \cos 3 \theta = 4 \sqrt{6}$

Multiply both sides by $\frac{3}{64}$ to get:

$\cos 3 \theta = \frac{3}{16} \sqrt{6}$

So:

$3 \theta = \pm {\cos}^{- 1} \left(\frac{3}{16} \sqrt{6}\right) + 2 k \pi$

So:

$\theta = \pm \frac{1}{3} {\cos}^{- 1} \left(\frac{3}{16} \sqrt{6}\right) + \frac{2 k \pi}{3}$

So:

$\cos \theta = \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{3}{16} \sqrt{6}\right) + \frac{2 k \pi}{3}\right)$

This takes three distinct values, for which we can use $k = 0 , 1 , 2$.

Hence zeros of our original cubic:

${x}_{k} = \frac{4}{3} \sqrt{6} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{3}{16} \sqrt{6}\right) + \frac{2 k \pi}{3}\right) \text{ } k = 0 , 1 , 2$

${x}_{0} \approx 3.051374241731$

${x}_{1} \approx - 2.534070196723$

${x}_{2} \approx - 0.517304045008$