#x^6 - 5x^3 + 8#................(factorise)?
2 Answers
#(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 = -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
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.
#= (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
#(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
Hence
#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
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
#x^3-3qx-p#
So if we can find three real zeros of this cubic, then we have the factorisation of the sextic