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.