# How do you factor x^6+5x^3+8?

Oct 28, 2016

${x}^{6} + 5 {x}^{3} + 8 = {\prod}_{n = 0}^{2} \left({x}^{2} - 2 \sqrt{2} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(- \frac{5 \sqrt{2}}{8}\right) + \frac{2 n \pi}{3}\right) x + 2\right)$

#### Explanation:

$f \left(x\right) = {x}^{6} + 5 {x}^{3} + 8$

The most direct way to factor this sextic polynomial is to factor it as a quadratic in ${x}^{3}$ first.

The downside of this approach is that it immediately requires some Complex arithmetic, but let's give it a go...

${x}^{6} + 5 {x}^{3} + 8 = {\left({x}^{3}\right)}^{2} + 5 \left({x}^{3}\right) + \frac{25}{4} + \frac{7}{4}$

$\textcolor{w h i t e}{{x}^{6} + 5 {x}^{3} + 8} = {\left({x}^{3} + \frac{5}{2}\right)}^{2} + {\left(\frac{\sqrt{7}}{2}\right)}^{2}$

$\textcolor{w h i t e}{{x}^{6} + 5 {x}^{3} + 8} = {\left({x}^{3} + \frac{5}{2}\right)}^{2} - {\left(\frac{\sqrt{7}}{2} i\right)}^{2}$

$\textcolor{w h i t e}{{x}^{6} + 5 {x}^{3} + 8} = \left(\left({x}^{3} + \frac{5}{2}\right) - \frac{\sqrt{7}}{2} i\right) \left(\left({x}^{3} + \frac{5}{2}\right) + \frac{\sqrt{7}}{2} i\right)$

$\textcolor{w h i t e}{{x}^{6} + 5 {x}^{3} + 8} = \left({x}^{3} + \frac{5}{2} - \frac{\sqrt{7}}{2} i\right) \left({x}^{3} + \frac{5}{2} + \frac{\sqrt{7}}{2} i\right)$

This has zeros when:

${x}^{3} = - \frac{5}{2} \pm \frac{\sqrt{7}}{2} i$

$\left\mid - \frac{5}{2} \pm \frac{\sqrt{7}}{2} i \right\mid = \sqrt{{\left(\frac{5}{2}\right)}^{2} + {\left(\frac{\sqrt{7}}{2}\right)}^{2}} = \sqrt{\frac{25}{4} + \frac{7}{4}} = \sqrt{\frac{32}{4}} = \sqrt{8} = 2 \sqrt{2}$

So:

${x}^{3} = 2 \sqrt{2} \left(\cos \theta \pm i \sin \theta\right)$

where $\theta = {\cos}^{- 1} \left(- \frac{5 \sqrt{2}}{8}\right)$

Hence:

$x = \sqrt{2} \left(\cos \left(\frac{\theta}{3} + \frac{2 n \pi}{3}\right) \pm i \sin \left(\frac{\theta}{3} + \frac{2 n \pi}{3}\right)\right)$

The quadratic factors with Real coefficients can be found by multiplying the corresponding linear factors in Complex conjugate pairs, e.g.:

(x-sqrt(2)(cos (theta/3+(2npi)/3) + i sin (theta/3+(2npi)/3))(x-sqrt(2)(cos (theta/3+(2npi)/3) - i sin (theta/3+(2npi)/3))

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

So:

${x}^{6} + 5 {x}^{3} + 8 = {\prod}_{n = 0}^{2} \left({x}^{2} - 2 \sqrt{2} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(- \frac{5 \sqrt{2}}{8}\right) + \frac{2 n \pi}{3}\right) x + 2\right)$

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Footnote

I am not happy with this answer, because I know that this sextic has a simple quadratic factor with integer coefficients and the full factorisation does not require trigonometric functions to express the coefficients.

I may produce a better answer later.

Oct 29, 2016

${x}^{6} + 5 {x}^{3} + 8 = \left({x}^{2} - x + 2\right) \left({x}^{4} + {x}^{3} - {x}^{2} + 2 x + 4\right)$

#### Explanation:

${x}^{6} + 5 {x}^{3} + 8$

Use a numerical method (Durand-Kerner) to find approximate zeros:

${x}_{1 , 2} \approx 0.5 \pm 1.32288 i$

${x}_{3 , 4} \approx 0.895644 \pm 1.09445 i$

${x}_{5 , 6} \approx - 1.39564 \pm 0.228425 i$

Taking these zeros in Complex conjugate pairs, we can find approximations for the quadratic factors with Real coefficients of ${x}^{6} + 5 {x}^{3} + 8$.

Note in particular that the Real part of ${x}_{1 , 2}$ is $\frac{1}{2}$, so let us find out what $\left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$ is:

$\left(x - 0.5 - 1.32288 i\right) \left(x - 0.5 + 1.32288 i\right) = {x}^{2} - x + 2.0000114944$

That looks supiciously like ${x}^{2} - x + 2$.

Next divide ${x}^{6} + 5 {x}^{3} + 8$ by ${x}^{2} - x + 2$ by long dividing the coefficients (not forgetting to include $0$'s for any missing powers of $x$... So:

${x}^{6} + 5 {x}^{3} + 8 = \left({x}^{2} - x + 2\right) \left({x}^{4} + {x}^{3} - {x}^{2} + 2 x + 4\right)$

Note that $\left(x - {x}_{3}\right) \left(x - {x}_{4}\right)$ and $\left(x - {x}_{5}\right) \left(x - {x}_{6}\right)$ will not give us integer coefficients, so this is a complete factorisation for integer coefficients.

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Footnote

There's some more description of the Durand-Kerner method at https://socratic.org/s/aza5vnRm

Here's the C++ program implementing the Durand Kerner algorithm I used to find the numerical approximations for this sextic... Nov 1, 2016

Yet another approach. See explanation

#### Explanation:

Like GPS taking us to the destination by different routes, I am trying to reach where respected George C is, by using a different approach.

The signs of the coefficients in $f \left(x\right) = {x}^{6} + 5 {x}^{3} + 8 \mathmr{and} f \left(- x\right) = {x}^{6} - {x}^{3} + 8$ reveal that either all linear factors of f(x) are complex, or four are complex and two are real.

Now,
$f \left(r {e}^{i \theta}\right) = a + i b$, where r > 0,
$a = {r}^{6} \cos 6 \theta + 5 {r}^{3} \cos 3 \theta$ and
$b = {r}^{6} \sin 6 \theta + 5 {r}^{3} \sin 3 \theta$.
$f = 0 \to a = 0 = b$.

$b = 0$ gives:

${r}^{3} \left({r}^{3} \sin 6 \theta + 5 \sin 3 \theta\right)$
$= {r}^{3} \sin 3 \theta \left(2 {r}^{3} \cos 3 \theta + 5\right)$
$= 0 \text{ }$ And so,
$\sin 3 \theta = 0 \to 3 \theta = k \pi \to \theta = k \frac{\pi}{3} , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$
and
$\cos 3 \theta = - \frac{5}{2 {r}^{3}} \text{ in } {Q}_{2} \mathmr{and} {Q}_{3}$

For $\theta = k \frac{\pi}{3} , a = 0$ gives
${r}^{6} - 5 {r}^{3} + 8 = 0 \to {r}^{3} = u n r e a l \frac{5 \pm \sqrt{25 - 32}}{2}$. So, ignored.
For $\cos 3 \theta = - \frac{5}{2 {r}^{3}} , b = 0$ gives (after simplification )
${r}^{6} = 8 \to r = \sqrt{2}$.

So, all the roots have the same modulus r =$\sqrt{2}$

Corresponding $\cos 3 \theta = - \frac{5}{4 \sqrt{2}} \text{ in } {Q}_{2} \mathmr{and} {Q}_{3.}$

The general value
$3 \theta = 2 k \pi + {\cos}^{- 1} \left(- \frac{5}{4} \sqrt{2}\right) = 350 {k}^{o} + {152.1144}^{o}$. So,
$\theta - 120 {k}^{\oplus} {50.7081}^{o}$

Corresponding cyclic values of $\cos \theta$

$\left(\cos {\theta}_{1} , \cos {\theta}_{2} , \cos {\theta}_{3}\right) = \left(0.633316 , 0.353553 , - 0.98587\right)$ for
$\left({\theta}_{1} , {\theta}_{2} , {\theta}_{3}\right) = \left({50.7048}^{o} , {290.705}^{o} , {530.705}^{o}\right)$, respectively.

The six points {(sqrt 2, +-theta_i}, i=1,2,3 lie on the circle $r = \sqrt{2}$ in the z-plane.

The directions of position vectors to these points are given by
$\left(\pm {50.7048}^{o} , \pm {69.295}^{o} , \pm {170.705}^{o}\right)$, respectively.

Complex roots occur in conjugate pairs and the quadratic factors for the pairs are
$\left({x}^{2} - 2 r \cos {\theta}_{1} + {r}^{2}\right) \left({x}^{2} - 2 r \cos {\theta}_{2} + {r}^{2}\right) \left({x}^{2} - 2 r \cos {\theta}_{3} + {r}^{2}\right)$

This is wholly tallying with the form already presented by George C.