To solve the problem of finding a polynomial function with real coefficients given the zeros $2$ and $-3i$, here is the detailed approach:

Key Steps:

1. Understand Real Coefficient Property:

   • If a polynomial has real coefficients and a complex root exists, its conjugate must also be a root.
   • Given $-3i$, its conjugate $3i$ is also a root.

2. List all Zeros:

   • The zeros of the polynomial are $2$, $-3i$, and $3i$.

3. Express Polynomial from Zeros:

   • The polynomial can be written as a product of factors:
     $f(x) = (x - r_1)(x - r_2)(x - r_3)$ where $r_1, r_2, r_3$ are the roots $2$, $-3i$, and $3i$.

4. Substitute the Roots:

   • $f(x) = (x - 2)(x - (-3i))(x - 3i)$
   • Simplify:
     $f(x) = (x - 2)(x + 3i)(x - 3i)$

5. Simplify the Complex Factors:

   • Multiply the complex conjugates $(x + 3i)(x - 3i)$:
     $(x + 3i)(x - 3i) = x^2 - (3i)^2 = x^2 - (-9) = x^2 + 9$
   • Therefore:
     $f(x) = (x - 2)(x^2 + 9)$

6. Expand the Expression:

   • Distribute $(x - 2)$ over $(x^2 + 9)$:
     $f(x) = x(x^2 + 9) - 2(x^2 + 9)$ Simplify each term:
     $f(x) = x^3 + 9x - 2x^2 - 18$
   • Rearrange into standard form:
     $f(x) = x^3 - 2x^2 + 9x - 18$

Final Polynomial:

The polynomial function is:
$f(x) = x^3 - 2x^2 + 9x - 18$

Do you need further explanation or have any follow-up questions? Let me know!

Related Questions:

1. How do you find a polynomial if more roots (e.g., $1 \pm i$) are given?
2. What are the properties of complex roots in polynomials with real coefficients?
3. How do you factorize a cubic polynomial like $x^3 - 2x^2 + 9x - 18$?
4. How can you check if a given polynomial has real coefficients?
5. How do conjugates apply to higher-degree polynomials?

Tip:

Always remember that complex roots appear in conjugate pairs when dealing with polynomials that have real coefficients.