Question 390322
By Vieta's formulas, the sum of the roots is real, so the sum of the other roots must be C - 6i for real C. There are actually infinitely many polynomials satisfying this.


We know the polynomial must be in the form


{{{P(x) = (x-i)(x-5i)Q(x)}}} where Q(x) is a polynomial with complex coefficients. If we let two roots of Q(x) be -5i and -i, then the polynomial will have real coefficients. Therefore we can let {{{Q(x) = (x+i)(x+5i)R(x)}}} where R(x) has real coefficients.


{{{P(x) = (x-i)(x-5i)(x+i)(x+5i)R(x) = (x-i)(x+i)(x-5i)(x+5i)R(x) = (x^2 + 1)(x^2 + 25)R(x) = (x^4 + 26x^2 + 25)R(x)}}} where R(x) is a polynomial with real coefficients. I'm sure many other solutions exist, but the problem only asks for one polynomial.