To find the polynomial $f(x)$ of degree 3 with integer coefficients, given the zeros $8i$ and $\frac{6}{5}$, we can proceed as follows:

Step-by-Step Solution:

1. Complex Conjugate Root Theorem: If a polynomial has real coefficients, then any non-real roots must come in conjugate pairs. Since $8i$ is a zero, its conjugate $-8i$ must also be a zero.

2. List the zeros: The zeros are:

   • $8i$
   • $-8i$
   • $\frac{6}{5}$

3. Form the corresponding factors:

   • For $8i$ and $-8i$: The corresponding quadratic factor is $(x - 8i)(x + 8i) = x^2 + 64$.
   • For $\frac{6}{5}$: The corresponding linear factor is $(x - \frac{6}{5})$. To make the coefficients integer, we can multiply this factor by 5, giving $(5x - 6)$.

4. Combine all factors: $f(x) = (x^2 + 64)(5x - 6)$

5. Expand the expression: $f(x) = 5x^3 - 6x^2 + 320x - 384$

Thus, the polynomial that satisfies the conditions is: $f(x) = 5x^3 - 6x^2 + 320x - 384$

Would you like more details on any step, or do you have any other questions?

Related Questions:

1. What is the importance of the Complex Conjugate Root Theorem?
2. How can you verify that a polynomial has integer coefficients?
3. What are other ways to factor polynomials if the roots are fractions?
4. Why must polynomials with real coefficients have complex roots in conjugate pairs?
5. Can the polynomial be factored further, or is it already in its simplest form?

Tip:

When working with polynomials with real coefficients, always remember that if there is a non-real root (like $8i$), its complex conjugate (like $-8i$) must also be a root.