# 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

#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

The sum of cubes identity can be written:

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

Use this with

#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

#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