Verified answer

They are both 2.

there is yet another answer very very corresponding to Dragan's that's won through adjusting the indications of the x^a million phrases in his factorization yielding: (x^3+2*ok^2*x^2-2*ok*x+a million) * (x^3-2*ok^2*x^2-2*ok*x-a million) = x^6 - 4*ok*(ok^3+a million)*x^4 - a million which provides the series n= a million,9,40 two .. Dragan's series shifted through ok @mein Hoon na for n=0 -a million,a million, +/-j and +/-j^2 are all roots. the completed factorization over the integers is then (x-a million)(x+a million)(x^2+x+a million)(x^2-x+a million) and you are able to combine issues into 3+3 or 4+2 or 5+a million at will @Gianlino you additionally could have 3 polynomials of degree 2 for n=0