# How do you factor completely F(x) = 2x^3 - 7x + 1?

Mar 29, 2017

Find the zeros of $F \left(x\right)$ to find:

$F \left(x\right) = 2 \left(x - {x}_{0}\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

where:

${x}_{n} = \frac{\sqrt{42}}{3} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(- \frac{3 \sqrt{42}}{98}\right) + \frac{2 n \pi}{3}\right)$

#### Explanation:

This cubic function has three irrational real zeros, for which we can find expressions using a trigonometric substitution...

Let $x = k \cos \theta$

Then:

$F \left(x\right) = F \left(k \cos \theta\right) = 2 {k}^{3} {\cos}^{3} \theta - 7 k \cos \theta + 1$

I would like this to contain something like:

$4 {\cos}^{3} \theta - 3 \cos \theta = \cos 3 \theta$

So I would like:

$\frac{2 {k}^{3}}{7 k} = \frac{4}{3}$

Choose $k = \frac{\sqrt{42}}{3}$

Then:

$\frac{2 {k}^{3}}{4} = \frac{7 \sqrt{42}}{9}$

So:

$2 {k}^{3} {\cos}^{3} \theta - 7 k \cos \theta + 1 = \frac{7 \sqrt{42}}{9} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) + 1$

$\textcolor{w h i t e}{2 {k}^{3} {\cos}^{3} \theta - 7 k \cos \theta + 1} = \frac{7 \sqrt{42}}{9} \cos 3 \theta + 1$

This has zeros when:

$\cos 3 \theta = - \frac{9}{7 \sqrt{42}} = - \frac{3 \sqrt{42}}{98}$

That is when:

$3 \theta = \pm \left({\cos}^{- 1} \left(- \frac{3 \sqrt{42}}{98}\right) + 2 n \pi\right) \text{ }$ for integer values of $n$

So:

$\cos \theta = \cos \left(\frac{1}{3} \left(\pm {\cos}^{- 1} \left(- \frac{3 \sqrt{42}}{98}\right) + 2 n \pi\right)\right)$

Hence using $x = k \cos \theta$ we find distinct zeros of $F \left(x\right)$:

${x}_{n} = \frac{\sqrt{42}}{3} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(- \frac{3 \sqrt{42}}{98}\right) + \frac{2 n \pi}{3}\right)$

for $n = 0 , 1 , 2$

Hence:

$F \left(x\right) = 2 \left(x - {x}_{0}\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

where:

${x}_{n} = \frac{\sqrt{42}}{3} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(- \frac{3 \sqrt{42}}{98}\right) + \frac{2 n \pi}{3}\right)$