If 2 + 5i is a zero, then by the complex conjugate root theorem, we must have its conjugate as a zero to have a polynomial containing real coefficients. Therefore, the zeros are -3, 2 + 5i, and 2 - 5i. We have three zeros so this is a degree 3 polynomial (n = 3).
f(x) has the equation
f(x) = (x+3)(x - (2 + 5i))(x - (2 - 5i))
If we expand this polynomial out, we get the simplest standard form
f(x) = x^3-x^2+17x+87
Therefore the answer to this question is f(x) = x^3-x^2+17x+87
9(2k+3)+2=11 (k-y)
k=-11/7 y+ -29/7
My guess would be x = -30
