The problem asks to find a polynomial $f(x)$ of the lowest degree with integer coefficients and the given complex zeros $-3i$ and $4 + 3i$.

Key Concepts:

• Complex Conjugates: If a polynomial has real coefficients, any complex roots must come in conjugate pairs.
  • Given roots: $-3i$ and $4 + 3i$
  • Their conjugates: $3i$ and $4 - 3i$

Steps:

1. The zeros of the polynomial are:

   • $-3i$, $3i$, $4 + 3i$, and $4 - 3i$.

2. The corresponding factors of $f(x)$ for these zeros are:

   • $(x + 3i)$ and $(x - 3i)$ for $-3i$ and $3i$,
   • $(x - (4 + 3i))$ and $(x - (4 - 3i))$ for $4 + 3i$ and $4 - 3i$.

3. To form the polynomial:

   • Multiply the factors for the imaginary part: $(x + 3i)(x - 3i) = x^2 + 9$
   • Multiply the factors for the real part: $(x - (4 + 3i))(x - (4 - 3i)) = \left( x - 4 \right)^2 - (3i)^2 = (x - 4)^2 + 9 = x^2 - 8x + 25$

4. Now, multiply these two quadratic expressions: $f(x) = (x^2 + 9)(x^2 - 8x + 25)$ Expand this: $f(x) = x^4 - 8x^3 + 25x^2 + 9x^2 - 72x + 225$ Simplify: $f(x) = x^4 - 8x^3 + 34x^2 - 72x + 225$

Final Answer:

The polynomial is: $f(x) = x^4 - 8x^3 + 34x^2 - 72x + 225$

Let me know if you'd like further details or have any questions!

Here are 5 related questions to expand on this topic:

1. What are complex conjugates, and why must they come in pairs for polynomials with real coefficients?
2. How would the polynomial change if the roots were real instead of complex?
3. Can we verify the result by substituting the zeros back into the polynomial?
4. What would happen if one of the complex conjugate pairs was missing from the polynomial?
5. How can the degree of a polynomial help determine the number of zeros it has?

Tip: When expanding products of polynomials, it helps to group like terms to simplify calculations quickly.