The problem asks for a polynomial $f(x)$ of degree 3 with real coefficients, given the zeros $4$, $-2 + 3i$, and its conjugate $-2 - 3i$. Since the coefficients are real, complex roots must occur in conjugate pairs, which is why $-2 - 3i$ is included.

To find the polynomial, we can use the fact that the polynomial is the product of factors corresponding to its roots:

$f(x) = (x - 4)(x - (-2 + 3i))(x - (-2 - 3i))$

This simplifies to:

$f(x) = (x - 4)((x + 2) - 3i)((x + 2) + 3i)$

Now, simplify the product of the complex conjugates:

$((x + 2) - 3i)((x + 2) + 3i) = (x + 2)^2 - (3i)^2 = (x + 2)^2 + 9$

Expanding $(x + 2)^2$:

$(x + 2)^2 = x^2 + 4x + 4$

Thus, the product of the complex conjugates is:

$(x^2 + 4x + 4) + 9 = x^2 + 4x + 13$

Now, multiply this by $(x - 4)$:

$f(x) = (x - 4)(x^2 + 4x + 13)$

Expand this expression:

$f(x) = x(x^2 + 4x + 13) - 4(x^2 + 4x + 13)$ $f(x) = x^3 + 4x^2 + 13x - 4x^2 - 16x - 52$ $f(x) = x^3 - 3x - 52$

Thus, the polynomial is:

$f(x) = x^3 - 3x - 52$

This should be the correct answer.

Would you like further clarification or details on this solution?

Here are 5 follow-up questions related to this topic:

1. How do you identify the conjugate of a complex number?
2. Why must complex roots of polynomials with real coefficients come in conjugate pairs?
3. What are the general steps to expand and simplify polynomial expressions?
4. How would the polynomial change if a different set of roots were given?
5. Can you explain the role of each term in the final polynomial expression?

Tip: Always remember that for polynomials with real coefficients, every non-real complex root must have a corresponding conjugate root.