To solve this problem, we need to construct a third-degree polynomial function $f(x)$ with the given conditions:

1. Degree: $n = 3$
2. Zeros: $4$ and $5i$ are zeros.
3. Function value: $f(1) = -78$.

Step 1: Determine All Zeros

Since the polynomial has real coefficients, the complex zero $5i$ implies that $-5i$ is also a zero. Therefore, the three zeros are: $4, \quad 5i, \quad \text{and} \quad -5i.$

Step 2: Write the Polynomial in Factored Form

Using these zeros, we can express $f(x)$ as: $f(x) = a(x - 4)(x - 5i)(x + 5i),$ where $a$ is a constant we will determine.

Step 3: Simplify the Expression

Notice that $(x - 5i)(x + 5i)$ is a difference of squares, which simplifies to: $(x - 5i)(x + 5i) = x^2 + 25.$ So, we have: $f(x) = a(x - 4)(x^2 + 25).$

Step 4: Expand the Polynomial

Now, expand $f(x)$: $f(x) = a(x^3 + 25x - 4x^2 - 100).$ Rearrange terms to get: $f(x) = a(x^3 - 4x^2 + 25x - 100).$

Step 5: Use $f(1) = -78$ to Find $a$

Substitute $x = 1$ into $f(x)$ and set it equal to $-78$: $f(1) = a(1^3 - 4 \cdot 1^2 + 25 \cdot 1 - 100) = -78.$ Simplify inside the parentheses: $f(1) = a(1 - 4 + 25 - 100) = -78,$ $f(1) = a(-78).$ Now, solve for $a$: $a(-78) = -78,$ $a = 1.$

Final Polynomial

Thus, the polynomial is: $f(x) = x^3 - 4x^2 + 25x - 100.$

Would you like a deeper breakdown of any steps or additional practice problems on polynomial functions?

Here are some questions that could further enhance your understanding:

1. What would the polynomial look like if the zeros were $4$, $5i$, and $-3$ instead?
2. How does the presence of complex roots affect the coefficients of a polynomial with real coefficients?
3. What is the difference between finding real and complex roots in polynomial functions?
4. How would we change the function if we needed $f(2) = -100$ instead of $f(1) = -78$?
5. How do you determine the degree of a polynomial based on its number of zeros?

Tip: Complex roots in polynomials with real coefficients always come in conjugate pairs, such as $a + bi$ and $a - bi$.