Question

How do you find a polynomial function of lowest degree with rational coefficients that has the given number of some of it's zeros. -5i, 3?

Answer 1

`f(x) = (x-5i)(x+5i)(x-3) = x^3-3x^2+25x-75`

Explanation:

If the coefficients are real (let alone rational), then any complex zeros will occur in conjugate pairs.

So the roots of `f(x) = 0` are at least `+-5i` and `3`.

Hence

`f(x) = (x-5i)(x+5i)(x-3)`

`= (x^2+25)(x-3)= x^3-3x^2+25x-75`

Any polynomial in `x` with these zeros will be a multiple of `f(x)`