How do you factor #13(x^6+ 1)^4(18x^5)(9x + 2)^3 + 9(9x + 2)^2(9)(x^6 + 1)^5#?

1 Answer
Apr 15, 2016

#13(x^6+1)^4(18x^5)(9x+2)^3+9(9x+2)^2(9)(x^6+1)^5#

#=9(x^2+1)^4(x^2-sqrt(3)x+1)^4(x^2+sqrt(3)x+1)^4(9x+2)^2(243x^6+52x^5+9)#

Explanation:

Given:

#13(x^6+1)^4(18x^5)(9x+2)^3+9(9x+2)^2(9)(x^6+1)^5#

First note that both of the terms are divisible by:

#9(x^6+1)^4(9x+2)^2#

So separate that out as a factor first to get:

#13(x^6+1)^4(18x^5)(9x+2)^3+9(9x+2)^2(9)(x^6+1)^5#

#=9(x^6+1)^4(9x+2)^2(13(2x^5)(9x+2)+(9)(x^6+1))#

#=9(x^6+1)^4(9x+2)^2(243x^6+52x^5+9)#

We can treat #x^6+1# as a sum of cubes to factorise it some way:

The sum of cubes identity can be written:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

Use this with #a=x^2# and #b=1# to find:

#x^6+1 = (x^2)^3+1^3 = (x^2+1)(x^4-x^2+1)#

Next note that:

#(a^2-kab+b^2)(a^2+kab+b^2)=(a^4+(2-k^2)a^2b^2+b^4)#

So, putting #a=x#, #b=1# and #k=sqrt(3)# we find:

#x^4-x^2+1 = (x^2-sqrt(3)x+1)(x^2+sqrt(3)x+1)#

So:

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

Putting it all together:

#13(x^6+1)^4(18x^5)(9x+2)^3+9(9x+2)^2(9)(x^6+1)^5#

#=9(x^2+1)^4(x^2-sqrt(3)x+1)^4(x^2+sqrt(3)x+1)^4(9x+2)^2(243x^6+52x^5+9)#

The remaining quadratic factors have no Real zeros and therefore no linear factors with Real coefficients.

The remaining sextic factor has #6# Complex zeros in #3# conjugate pairs. That means that in theory it could be factored into #3# quadratic factors with Real coefficients, but as far as I can tell the coefficients of those factors are not expressible in terms of #n#th roots and other ordinary arithmetic operations. That is, they are not determinable by practical algebraic means. It is possible to find numeric approximations using Newton Raphson, Durand-Kerner or other similar numeric methods.