Find an nth-degree polynomial function with real coefficients satisfying the given conditions If you are using a graphing utility, use it to graph the function and verify the real zeros and the given function value. \[ \begin{array}{l} \mathrm{n}=3 \\ 2 \text { and } 4 i \text { are zeros; } \\ \mathrm{f}(-1)=102 \end{array} \] \[ f(x)= \]

Step 1 ：The problem is asking for a polynomial of degree 3 that has the roots 2 and 4i. Since the coefficients are real, the complex roots must come in conjugate pairs. Therefore, the other root is -4i. The polynomial can be written in factored form as: \(f(x) = a*(x - 2)(x - 4i)(x + 4i)\)

Step 2 ：We can simplify this to: \(f(x) = a*(x - 2)(x^2 + 16)\)

Step 3 ：We also know that \(f(-1) = 102\). We can substitute \(x = -1\) into the equation to solve for a: \(102 = a*(-1 - 2)*((-1)^2 + 16)\)

Step 4 ：Solving the above equation, we find that \(a = -2\)

Step 5 ：Now we can substitute \(a = -2\) into the polynomial to get the final polynomial function: \(f(x) = -2(x - 2)(x^2 + 16)\)

Step 6 ：Finally, simplifying the polynomial function, we get \(\boxed{f(x) = -2x^3 + 4x^2 + 32x - 64}\)