Find an nth-degree polynomial function with real coefficients satisfying the \[ \begin{array}{l} n=3 \\ 4 \text { and } 5 i \text { are zeros; } \\ f(2)=-174 \end{array} \] $f(x)=$ (Type an expression using $x$ as the variable. Simplify your answer.)

Step 1 ：Given that the polynomial is of degree 3, it has 3 roots. The roots are 4, 5i, and -5i.

Step 2 ：The general form of a polynomial is given by: \(f(x) = a(x - r1)(x - r2)(x - r3)\), where \(r1, r2, r3\) are the roots of the polynomial and \(a\) is a constant.

Step 3 ：Substitute the roots into the equation to get the polynomial: \(f(x) = a(x - 4)(x - 5i)(x + 5i)\)

Step 4 ：We know that the polynomial passes through the point (2, -174). Substitute this point into the polynomial to solve for the constant \(a\).

Step 5 ：After solving, we find that \(a = 3.0\).

Step 6 ：Substitute \(a\) back into the polynomial to get the final polynomial function: \(f(x) = 3.0(x - 4)(x - 5i)(x + 5i)\)

Step 7 ：\(\boxed{f(x) = 3.0(x - 4)(x - 5i)(x + 5i)}\) is the final answer.