The problem here is to find an nth-degree polynomial function with real coefficients that satisfies the given conditions:

1. Degree of the polynomial, $n = 3$.
2. Zeros of the polynomial include $-3$ and $3 + 4i$.
3. A function value condition: $f(-1) = 64$.

Solution

1. Identify the zeros: Since the polynomial has real coefficients, the complex conjugate $3 - 4i$ must also be a zero. This gives us three zeros: $-3$, $3 + 4i$, and $3 - 4i$.

2. Write the polynomial in terms of its factors: Since the zeros are $-3$, $3 + 4i$, and $3 - 4i$, the polynomial can be written as: $f(x) = a(x + 3)(x - (3 + 4i))(x - (3 - 4i))$ where $a$ is a constant.

3. Expand the complex factors: The factors $(x - (3 + 4i))$ and $(x - (3 - 4i))$ can be combined: $(x - (3 + 4i))(x - (3 - 4i)) = (x - 3)^2 - (4i)^2$ Simplify by calculating $(4i)^2$: $= (x - 3)^2 - (-16) = (x - 3)^2 + 16$ Now expand $(x - 3)^2 + 16$: $= x^2 - 6x + 9 + 16 = x^2 - 6x + 25$

4. Form the polynomial: Substitute back to get: $f(x) = a(x + 3)(x^2 - 6x + 25)$

5. Use the condition $f(-1) = 64$ to find $a$: Substitute $x = -1$ into $f(x)$: $f(-1) = a(-1 + 3)((-1)^2 - 6(-1) + 25)$ Simplify each part: $= a(2)(1 + 6 + 25) = a \cdot 2 \cdot 32 = 64$ Solving for $a$: $64 = 64a \implies a = 1$

6. Final polynomial: Substitute $a = 1$ into $f(x)$: $f(x) = (x + 3)(x^2 - 6x + 25)$ Expand if needed: $f(x) = x^3 - 3x^2 + 7x + 75$

Final Answer

$f(x) = x^3 - 3x^2 + 7x + 75$

Would you like more details on any step? Here are some questions you might find helpful:

1. Why do complex roots appear in conjugate pairs for polynomials with real coefficients?
2. What is the significance of the constant $a$ in polynomial functions?
3. How would the solution change if there were no function value condition (like $f(-1) = 64$)?
4. Can we apply this method to higher degree polynomials with more complex roots?
5. What role does polynomial expansion play in this solution?

Tip: When you have complex roots, always remember to include the conjugate to ensure real coefficients in your polynomial.