The problem is asking you to write a degree 3 polynomial $f(x)$ that has integer coefficients and the given zeros: $-4i$ and $\frac{8}{5}$.

Key Points:

1. Since complex zeros always come in conjugate pairs for polynomials with real coefficients, the conjugate of $-4i$, which is $4i$, must also be a root.
2. The roots of the polynomial are:
   • $-4i$
   • $4i$
   • $\frac{8}{5}$

Steps to construct the polynomial:

1. The factors corresponding to these zeros are:

   • $(x + 4i)$ for $-4i$
   • $(x - 4i)$ for $4i$
   • $\left(x - \frac{8}{5}\right)$ for $\frac{8}{5}$

2. To ensure integer coefficients, multiply $\left(x - \frac{8}{5}\right)$ by 5: $5\left(x - \frac{8}{5}\right) = 5x - 8$

3. Multiply the conjugate complex factors: $(x + 4i)(x - 4i) = x^2 + 16$ (since $(a + bi)(a - bi) = a^2 + b^2$).

4. Now, the polynomial is: $f(x) = (x^2 + 16)(5x - 8)$

5. Expand the polynomial: $f(x) = (x^2 + 16)(5x - 8) = 5x^3 - 8x^2 + 80x - 128$

Thus, the polynomial is: $f(x) = 5x^3 - 8x^2 + 80x - 128$

Would you like more details or have any questions? Here are some related questions for practice:

1. What is the importance of conjugate pairs in polynomials with real coefficients?
2. How do you expand polynomials with complex numbers?
3. Can you derive a polynomial if only a fraction is given as a zero?
4. How does multiplying by a factor like 5 affect the polynomial?
5. What role does the degree of the polynomial play in determining its structure?

Tip: When dealing with fractions as roots, always multiply by the denominator to clear it and ensure integer coefficients.