Question

Write a polynomial in standard form with zeros -2 and 3i...

Answer 1

\[(x+2)(x-3i)(x+3i)\] is a start

Answer 2

the product of the last two factors is \(x^2+9\) so you remaining job is to multiply
\[(x+2)(x^2+9)\]

Answer 3

Thank you for taking the time to help me but I have one more question. Why did you use both -3i and 3i?

Answer 4

because if \(3i\) is a zero of a polynomial with integer coefficients, then so is its "conjugate"
\(-3i\)

Answer 5

you know this from the quadratic formula
it is the \(\pm\sqrt{b^2-4ac}\) part
if \(b^2-4ac\) is negative you get two complex roots, and one is the conjugate of the other

Answer 6

Oh... Thank you so much!