For a degree 3 polynomial with integer coefficients, the given zeros are $-3i$ and $\frac{9}{7}$. Here's how we construct the polynomial step by step:

1. Use the fact that the coefficients are integers: Since the polynomial has real coefficients, the complex root $-3i$ must have its conjugate $3i$ as a root as well. Therefore, the roots are: $-3i, \ 3i, \ \frac{9}{7}$

2. Form factors from the roots:

   • From the roots $-3i$ and $3i$, the corresponding factor is: $(x + 3i)(x - 3i) = x^2 + 9$
   • From the root $\frac{9}{7}$, the corresponding factor is: $\left( x - \frac{9}{7} \right)$ To get integer coefficients, multiply by 7 to eliminate the fraction: $7 \left( x - \frac{9}{7} \right) = 7x - 9$

3. Form the polynomial: Multiply the factors $(x^2 + 9)$ and $(7x - 9)$ to obtain the polynomial: $P(x) = (x^2 + 9)(7x - 9)$ Expand this expression: $P(x) = (x^2 + 9)(7x - 9) = x^2(7x - 9) + 9(7x - 9)$ Simplify: $P(x) = 7x^3 - 9x^2 + 63x - 81$

Thus, the degree 3 polynomial with integer coefficients is: $P(x) = 7x^3 - 9x^2 + 63x - 81$

Would you like further explanation, or do you have any questions about the process?

Here are 5 related questions to expand on this concept:

1. How do complex conjugates affect the formation of polynomials with real coefficients?
2. Can you derive a polynomial if only two roots (one real, one complex) are given?
3. What is the relationship between roots and factors of a polynomial?
4. How would the polynomial change if all the roots were rational numbers?
5. What if you were asked for a polynomial of degree 4 with one real and one complex root?

Tip: When dealing with polynomials with complex roots, always include the conjugate of the complex root to ensure the coefficients remain real.