Find an nth-degree polynomial function with real coefficients, a leading coefficient of 1 , and satisfying the given conditions. $n=4 ; i$ and $12 i$ are zeros
$f(x)=\square$ (Simplify your answer.)

Step 1 ：The problem asks for a polynomial of degree 4 with real coefficients and a leading coefficient of 1. It also states that \(i\) and \(12i\) are zeros of the polynomial.

Step 2 ：Since the coefficients are real, the complex roots must come in conjugate pairs. Therefore, the other two roots must be \(-i\) and \(-12i\).

Step 3 ：The polynomial can be found by multiplying the binomials formed by the roots.

Step 4 ：Let's calculate the polynomial. The roots are \(i\), \(-i\), \(12i\), and \(-12i\).

Step 5 ：The polynomial is \(x^4 + 145x^2 + 144\). This polynomial is of degree 4, has real coefficients, and a leading coefficient of 1, as required by the problem.

Step 6 ：Final Answer: \(f(x)=\boxed{x^4 + 145x^2 + 144}\)