Write the equation of a cubic polynomial function with real coefficients that has the following as zeros.

Question

3 and -2i are the zeroes for the equation

Answer 1

This type of problem tests your knowledge of the Complex Conjugate Root Theorem, which states:

If a polynomial with real coefficients has a complex number as a root,

then the conjugate of the complex number is also a root.

What that means here is that since -2i is a root, so is its conjugate, +2i. So now you have three roots: 3, -2i, and 2i. Now use the factored form of a polynomial:

P(x) = a·(x-3)(x-(-2i))(x-2i) = a·(x-3)(x+2i)(x-2i)

where a is a constant. In fact, every value of a you choose will have these same three roots, so there are an infinite number of polynomials with these roots. If the problem also gave you a particular point that the polynomial goes through, then you could use that point to find the one specific value of a. Without a point given, however, you are free to choose any value of a you'd like except a = 0. The simplest is a = 1, which gives:

P(x) = (x-3)(x+2i)(x-2i)

Multiply it out if your teacher wants the answer in standard form; otherwise it's OK to leave in factored form.