We are given that the polynomial has real coefficients and the zeros are $2$ and $-2i$. Since the coefficients are real, the complex conjugate of any non-real zero must also be a zero of the polynomial. Therefore, the zero $2i$ (the conjugate of $-2i$) must also be a root.

Thus, the polynomial has the following roots:

• $2$
• $-2i$
• $2i$

The polynomial can be formed by multiplying the factors corresponding to these roots:

$f(x) = (x - 2)(x - (-2i))(x - 2i)$

Simplifying step by step:

1. Combine the complex factors: $(x + 2i)(x - 2i)$

This is a difference of squares: $(x + 2i)(x - 2i) = x^2 - (2i)^2 = x^2 - (-4) = x^2 + 4$

2. Multiply the result with $(x - 2)$: $f(x) = (x - 2)(x^2 + 4)$

Now expand: $f(x) = x(x^2 + 4) - 2(x^2 + 4) = x^3 + 4x - 2x^2 - 8$

Thus, the polynomial function is: $f(x) = x^3 - 2x^2 + 4x - 8$

Would you like further details or clarifications?

Here are some related questions:

1. How do you handle polynomials with higher degree complex roots?
2. What is the importance of including conjugate pairs in polynomials with real coefficients?
3. How would the polynomial change if the zero at $x = 2$ had multiplicity 2?
4. How does factoring complex roots help simplify polynomial expressions?
5. Can a polynomial with real coefficients have only complex roots?

Tip: Always check for complex conjugates when working with polynomials with real coefficients to ensure the equation remains valid.