#x^6 - 5x^3 + 8#................(factorise)?

2 Answers
Sep 13, 2017

#x^6-5x^3+8 =#

#(x^2-(alpha+bar(alpha))x+2)(x^2-(omegaalpha+omega^2bar(alpha))x+2)(x^2-(omega^2alpha+omegabar(alpha))x+2)#

as described below...

Explanation:

Warning:

This answer may well be more advanced than you are expected to know.

Given:

#x^6-5x^3+8#

Use the quadratic formula to find zeros:

#x^3 = (5+-sqrt(5^2-4(1)(8)))/(2(1)) = (5+-sqrt(7)i)/2#

Let:

#alpha = root(3)((5+sqrt(7)i)/2)#

Then:

#bar(alpha) = root(3)((5-sqrt(7)i)/2)#

These are two of the complex zeros of the given sextic. The other four come from multiplying by powers of #omega#, the primitive complex cube root of #1#:

#omega = -1/2+sqrt(3)/2i#

Combining these complex zeros in complex conjugate pairs, we find the quadratic factors with real coefficients:

#(x^2-(alpha+bar(alpha))x+alphabar(alpha))#

#(x^2-(omegaalpha+omega^2bar(alpha))x+alphabar(alpha))#

#(x^2-(omega^2alpha+omegabar(alpha))x+alphabar(alpha))#

Note that the product of the constant terms must be #8#.

So we have:

#(alphabar(alpha))^3 = 8#

Hence we have:

#alphabar(alpha) = 2#

So:

#x^6-5x^3+8 =#

#(x^2-(alpha+bar(alpha))x+2)(x^2-(omegaalpha+omega^2bar(alpha))x+2)(x^2-(omega^2alpha+omegabar(alpha))x+2)#

Notes

It is possible to simplify and find:

#alpha+bar(alpha) = 1/2(1+sqrt(21))#

#omegaalpha+omega^2bar(alpha) = 1/2(1-sqrt(21))#

#omega^2alpha+omegabar(alpha) = -1#

but it is not (yet) clear to me how best to do this.

Sep 13, 2017

#x^6-5x^3+8#

#= (x^2+x+2)(x^2+(-1/2+sqrt(21)/2)x+2)(x^2+(-1/2-sqrt(21)/2)x+2)#

Explanation:

Here's a simpler method...

Given:

#x^6-5x^3+8#

Look for a factorisation of the form:

#x^6-5x^3+8#

#= (x^2+alphax+2)(x^2+betax+2)(x^2+gammax+2)#

#=x^6+(alpha+beta+gamma)x^5+(alphabeta+betagamma+gammaalpha+6)x^4+(2(alpha+beta+gamma)+alphabetagamma)x^3+(2(alphabeta+betagamma+gammaalpha)+12)x^2+4(alpha+beta+gamma)x+8#

Equating coefficients we find:

#{ (alpha+beta+gamma = 0), (alphabeta+betagamma+gammaalpha = -6), (alphabetagamma = -5) :}#

So #alpha, beta, gamma# are the zeros of the cubic:

#(x-alpha)(x-beta)(x-gamma)#

#=x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma#

#=x^3-6x+5#

Note that the sum of the coefficients of this cubic is #0#. That is #1-6+5 = 0#.

Hence #x=1# is a zero and #(x-1)# a factor:

#x^3-6x+5 = (x-1)(x^2+x-5)#

The zeros of the remaining quadratic can be found using the quadratic formula as:

#x = (-1+-sqrt(1^2-4(1)(-5)))/(2(1)) = 1/2(-1+-sqrt(21))#

So #{ alpha, beta, gamma } = { 1, -1/2+sqrt(21)/2, -1/2-sqrt(21)/2 }#

So:

#x^6-5x^3+8#

#= (x^2+x+2)(x^2+(-1/2+sqrt(21)/2)x+2)(x^2+(-1/2-sqrt(21)/2)x+2)#

Bonus

Can we generalise the above derivation?

#x^6+px^3+q^3#

#=(x^2+alphax+q)(x^2+betax+q)(x^2+gammax+q)#

#=x^6+(alpha+beta+gamma)x^5+(alphabeta+betagamma+gammaalpha+3q)x^4+(q(alpha+beta+gamma)+alphabetagamma)x^3+q(alphabeta+betagamma+gammaalpha+3q)x^2+q^2(alpha+beta+gamma)x+q^3#

Equating coefficients:

#{ (alpha+beta+gamma = 0), (alphabeta+betagamma+gammaalpha = -3q), (alphabetagamma=p) :}#

Hence #alpha, beta, gamma# are the zeros of:

#x^3-3qx-p#

So if we can find three real zeros of this cubic, then we have the factorisation of the sextic #x^6+px^3+q^3# into three quadratics with real coefficients.