Question

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

Answer 1

`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.

Answer 2

`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.