A degree 3 polynomial with integer coefficients must have conjugate pairs of complex roots if it includes any complex roots. Given that $4i$ is a zero, its conjugate $-4i$ must also be a zero.

Thus, the polynomial will have three roots:

1. $4i$
2. $-4i$
3. $\frac{8}{7}$

Now, let's construct the polynomial.

1. The factor corresponding to the root $4i$ is $(x - 4i)$.
2. The factor corresponding to the root $-4i$ is $(x + 4i)$.
3. The factor corresponding to the root $\frac{8}{7}$ is $\left(7x - 8\right)$ because we need integer coefficients.

First, combine the factors corresponding to $4i$ and $-4i$:

$(x - 4i)(x + 4i) = x^2 - (4i)^2 = x^2 - (-16) = x^2 + 16$

Now, the polynomial is:

$(x^2 + 16)(7x - 8)$

Let's expand this:

$(x^2 + 16)(7x - 8) = 7x^3 - 8x^2 + 112x - 128$

Thus, the degree 3 polynomial is:

$7x^3 - 8x^2 + 112x - 128$

Would you like more details or have any questions?

Here are some related questions you might find useful:

1. How do you find the conjugate of a complex root?
2. What is the reasoning behind multiplying polynomials with rational roots?
3. How can you verify if a polynomial has integer coefficients?
4. What are the conditions for a polynomial to have complex roots?
5. Can a polynomial of degree 3 have only real roots?

Tip: Whenever a polynomial has a complex root, its conjugate must also be a root if the polynomial has real (or integer) coefficients.