Could you have a polynomial with solutions 2i and 5i? If not explain why not. If yes, then construct the polynomial.

If you ever have a polynomial with a solution of bi, then it will also have a solution of -bi.  Imaginary solutions always come in pairs.

So, yes, you could have a polynomial with solutions 2i and 5i, as long as -2i and -5i are also solutions.

(x-2i)(x+2i)(x-5i)(x+5i) would be the most basic polynomial you could form.

(x-2i)(x+2i)(x-5i)(x+5i)  = (x^2+4)(x^2+25)

                                   = x^4 + 29x^2 + 100

So the equation would be x^4 + 29x^2 + 100 = 0.

Now, if the question was "only the solutions of 2i and 5i and no others," then the answer is no, for the previously stated reason.