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The problem asks to find a polynomial \( f(x) \) of degree 3 with real coefficients, given that it has the zeros \( -3 \) and \( 2i \).

### Key Points:
- Since the polynomial has real coefficients, the complex zeros must occur in conjugate pairs. So, if \( 2i \) is a zero, its conjugate \( -2i \) must also be a zero.
- The zeros of the polynomial are \( -3 \), \( 2i \), and \( -2i \).

### Polynomial Construction:
The general form of a polynomial is:
\[
f(x) = a(x - r_1)(x - r_2)(x - r_3)
\]
where \( r_1, r_2, r_3 \) are the roots of the polynomial.

For this problem, the roots are \( -3 \), \( 2i \), and \( -2i \). Therefore, the polynomial is:
\[
f(x) = a(x + 3)(x - 2i)(x + 2i)
\]

Now, simplify:
\[
(x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 + 4
\]
So, the polynomial becomes:
\[
f(x) = a(x + 3)(x^2 + 4)
\]
Expanding this:
\[
f(x) = a \left( x(x^2 + 4) + 3(x^2 + 4) \right) = a(x^3 + 4x + 3x^2 + 12)
\]
Thus:
\[
f(x) = a(x^3 + 3x^2 + 4x + 12)
\]

Since no specific value of \( a \) is given, we can assume \( a = 1 \), giving:
\[
f(x) = x^3 + 3x^2 + 4x + 12
\]

This is the required polynomial.

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Would you like further details or have any questions?

Here are 5 related questions you might find interesting:
1. How do we determine the conjugate of a complex number?
2. Why do complex roots come in conjugate pairs for polynomials with real coefficients?
3. How does the degree of a polynomial relate to the number of its roots?
4. Can a polynomial have more than two complex conjugate pairs of roots?
5. What is the process of factoring a polynomial with both real and complex roots?

**Tip**: When dealing with polynomials with real coefficients, always remember that non-real roots must occur in conjugate pairs to maintain real coefficients.

Find a polynomial f(x) of degree 3 with real coefficients and the following zeros: -3, 2i.

Find a degree 3 polynomial with real coefficients and the zeros -3 and 2i. This problem involves key algebraic concepts like polynomials and complex numbers, suitable for students in Grades 10-12.

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