Question

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

Answer 1

`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.