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

##### 3 Answers

#### Answer:

#### Explanation:

The most direct way to factor this sextic polynomial is to factor it as a quadratic in

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

This has zeros when:

#x^3 = -5/2+-sqrt(7)/2i#

#abs(-5/2+-sqrt(7)/2i) = sqrt((5/2)^2+(sqrt(7)/2)^2) = sqrt(25/4+7/4) = sqrt(32/4) = sqrt(8) = 2sqrt(2)#

So:

#x^3 = 2sqrt(2)(cos theta +- i sin theta)#

where

Hence:

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

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^2-2sqrt(2) cos (theta/3+(2npi)/3)x + 2#

So:

#x^6+5x^3+8 = prod_(n=0)^2 (x^2-2sqrt(2) cos (1/3cos^(-1)(-(5sqrt(2))/8)+(2npi)/3)x + 2)#

**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.

#### Answer:

#### Explanation:

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

#x_(1,2) ~~ 0.5 +- 1.32288i#

#x_(3,4) ~~ 0.895644+-1.09445i#

#x_(5,6) ~~ -1.39564+-0.228425i#

Taking these zeros in Complex conjugate pairs, we can find approximations for the quadratic factors with Real coefficients of

Note in particular that the Real part of

#(x-0.5-1.32288i)(x-0.5+1.32288i) = x^2-x +2.0000114944#

That looks supiciously like

Next divide

So:

#x^6+5x^3+8 = (x^2-x+2)(x^4+x^3-x^2+2x+4)#

Note that

**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...

#### Answer:

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

Now,

and

For

For

So, all the roots have the same modulus r =

Corresponding

The general value

Corresponding cyclic values of

The six points

The directions of position vectors to these points are given by

Complex roots occur in conjugate pairs and the quadratic factors for the pairs are

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