Write a polynomial \( f(x) \) that satisfies the given conditions. Degree 3 polynomial with integer coefficients with zeros \( 4 i \) and \( \frac{8}{7} \) \[ f(x)=\square \]

To find a degree 3 polynomial with integer coefficients having zeros of \( 4i \) and \( \frac{8}{7} \), we need to remember that complex roots come in conjugate pairs when coefficients are integers. Therefore, the roots of the polynomial are \( 4i, -4i, \) and \( \frac{8}{7} \). Starting with the factored form of the polynomial based on its roots: \[ f(x) = (x - 4i)(x + 4i) \left(x - \frac{8}{7}\right) \] The factors for the complex roots can be simplified: \[ (x - 4i)(x + 4i) = x^2 + 16 \] Now, we include the linear factor for the rational root: \[ f(x) = (x^2 + 16)\left(x - \frac{8}{7}\right) \] To eliminate the fraction, multiply through by 7: \[ f(x) = 7(x^2 + 16)\left(x - \frac{8}{7}\right) = 7(x^2 + 16)(7x - 8) \] This leads us to: \[ f(x) = 7(x^3 - \frac{8}{7}x^2 + 16x - \frac{128}{7}) = 7x^3 - 8x^2 + 112x - 128 \] Thus, the polynomial is: \[ \boxed{7x^3 - 8x^2 + 112x - 128} \]