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

#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.

#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...

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.