Question 1000695
.
If a polynomial with real coefficients has the zero  5+i,  then it has  5-i  as a zero too.


If a polynomial with real coefficients has the zero  1+4i,  then it has  1-4i  as a zero too.


Thus the polynomial of the minimal degree with real coefficients is


p(x) = (x-(5+i))*(x-(5-i))*(x-(1+4i))*(x-(1-4i)).


Now,  (x-(5+i))*(x-(5-i)) = {{{x^2 - 10x + 26}}},


(x-(1+4i))*(x-(1-4i)) = {{{x^2 - 2x + 17}}}.


Hence,


p(x) = {{{(x^2 - 10x + 26)}}} . {{{(x^2 - 2x + 17)}}}.


You can open parentheses if you want/you need.