#x^11+y^11=(x+y)(x^10-x^9y+x^8y^2-x^7y^3+x^6y^4-x^5y^5+x^4y^6-x^3y^7+x^2y^8-xy^9+y^10)#

First notice that if #y = -x# then #y^11 = (-x)^11 = -x^11#. So #(x+y)# is a factor. The other factor is easy to find, requiring alternating #+-1# coefficients to cause the intermediate terms to cancel out.

If we were allowed complex number coefficients, there would be a complete factorization as follows:

#x^11+y^11=(x+y)(x+tau y)(x+tau ^2y)(x+tau ^3y)(x+tau ^4y)(x+tau ^5y)(x+tau ^6y)(x+tau ^7y)(x+tau ^8y)(x+tau ^9y)(x+tau ^10y)#

where #tau# stands for the complex 11th root of 1.

Clearly, none of the factors apart from #(x+y)# has real coefficients, so there are definitely no other linear factors with rational coefficients.

I think that if any strict subset of the complex linear factors is multiplied together, we get a polynomial with at least one complex coefficient.

UPDATE:

Actually #tau^10 = bar tau#, #tau^9 = bar (tau^2)#, etc.

So #(x+tau)(x+tau^10) = (x^2+(tau + bar tau)x+1)# does have real coefficients.

#tau = cos ((2pi)/11) + i sin ((2pi)/11)#

and #tau + bar tau = 2 cos ((2pi)/11)#

etc.