Question 1198491
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Such function has the root of {{{sqrt(7)}}}, which is conjugate to {{{-sqrt(7)}}}
over the field of rational numbers, and the root 1-i, conjugate to 1+i 
over the field of real numbers.


So, the full list of the roots consists of 5 numbers {0, {{{sqrt(7)}}}, {{{-sqrt(7)}}}, 1+i, 1-i}.


Thus your polynomial is


    p(x) = {{{(x-0)*(x-sqrt(7))*(x+sqrt(7))*(x-(1+i))*(x-(1-i))}}} = 

         = {{{x*(x^2-7)*((x-1)-i)*((x-1)+i)}}} = {{{x*(x^2-7)*((x-1)^2+1)}}} = {{{x*(x^2-7)*(x^2 - 2x +2)}}}.


You can transform it to any other equivalent form.
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Solved.