Question

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

Answer 1

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

Explanation:

`f(x) = x^6+5x^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+5x^3+8 = (x^3)^2+5(x^3)+25/4+7/4`

`color(white)(x^6+5x^3+8) = (x^3+5/2)^2+(sqrt(7)/2)^2`

`color(white)(x^6+5x^3+8) = (x^3+5/2)^2-(sqrt(7)/2i)^2`

`color(white)(x^6+5x^3+8) = ((x^3+5/2)-sqrt(7)/2i)((x^3+5/2)+sqrt(7)/2i)`

`color(white)(x^6+5x^3+8) = (x^3+5/2-sqrt(7)/2i)(x^3+5/2+sqrt(7)/2i)`

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 `theta = cos^(-1)(-(5sqrt(2))/8)`

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)`

`color(white)()`
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 2

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

Explanation:

`x^6+5x^3+8`

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 `x^6+5x^3+8`.

Note in particular that the Real part of `x_(1,2)` is `1/2`, so let us find out what `(x-x_1)(x-x_2)` is:

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

That looks supiciously like `x^2-x+2`.

Next divide `x^6+5x^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+5x^3+8 = (x^2-x+2)(x^4+x^3-x^2+2x+4)`

Note that `(x-x_3)(x-x_4)` and `(x-x_5)(x-x_6)` will not give us integer coefficients, so this is a complete factorisation for integer coefficients.

`color(white)()`
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 3

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(x)=x^6+5x^3+8 and f(-x)=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(re^(itheta))=a+ib`, where r > 0,
`a=r^6 cos 6theta+5 r^3 cos 3theta` and
`b=r^6 sin 6theta + 5 r^3 sin 3theta`.
`f=0 to a=0=b`.

`b=0` gives:

`r^3 ( r^3sin 6theta + 5 sin 3theta)`
`=r^3 sin 3theta ( 2 r^3 cos 3theta+5)`
`=0" "` And so,
`sin 3theta = 0 to 3theta=kpi to theta=kpi/3, k=0, +-1, +-2, +-3, ...`
and
`cos 3theta =-5/(2r^3) " in " Q_2 or Q_3`

For `theta = kpi/3, a=0` gives
`r^6-5r^3+8=0 to r^3= unreal (5+-sqrt(25-32))/2`. So, ignored.
For `cos 3theta =-5/(2r^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 3theta=-5/(4sqrt 2) " in " Q_2 and Q_3.`

The general value
`3theta=2kpi+cos^(-1)(-5/4sqrt 2)=350k^o +152.1144^o`. So,
`theta-120k^o+50.7081^o`

Corresponding cyclic values of `cos theta`

`(cos theta_1, cos theta_2, cos theta_3)=(0.633316, 0.353553, -0.98587)` for
`(theta_1, theta_2, theta_3)=(50.7048^o, 290.705^o, 530.705^o)`, 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
`(+-50.7048^o, +-69.295^o, +-170.705^o)`, respectively.

Complex roots occur in conjugate pairs and the quadratic factors for the pairs are
`(x^2-2r cos theta_1+r^2)(x^2-2r cos theta_2+r^2)(x^2-2r cos theta_3+r^2)`

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