How do you write a polynomial function of least degree with integral coefficients that has the given zeros?

Question

I'm stumped on a homework problem:

Write a polynomial function of least degree with integral coefficients that has the given zeros

3 + 2i, 3i

Answer 1

There are actually four solutions (zeros) because of conjugate pairs.

x = ±3i, 3 ± 2i

(x - 3i)(x + 3i) = x² + 3ix - 3ix - 9i² = x² - 9(-1) = x² + 9

[x - (3 - 2i)][x - (3 + 2i)] = x² - x(3 + 2i) - x(3 - 2i) + (3 - 2i)(3 + 2i)

= x² - 3x - 2ix - 3x + 2ix + (9 - 4i²)

= x² - 6x + 9 - 4(-1)

= x² - 6x + 13

x² + 9 + x² - 6x + 13 = 2x² - 6x + 22