Use the given zero to find the remaining zeros of the function. \[ h(x)=3 x^{4}+8 x^{3}+45 x^{2}+128 x-48 ; \text { zero: }-4 i \]

Step 1 ：The given function is a polynomial of degree 4, which means it will have 4 zeros. If a polynomial has a complex zero, then its conjugate is also a zero of the polynomial. Therefore, if -4i is a zero, then its conjugate, 4i, is also a zero.

Step 2 ：We can then perform synthetic division twice, once with -4i and once with 4i, to find the remaining zeros.

Step 3 ：Performing synthetic division with -4i, we get the polynomial \(3x^3 + x^2(8 - 12i) + x(-3 - 32i) + 12i\).

Step 4 ：Performing synthetic division with 4i, we get the polynomial \(3x^2 + 8x - 3\).

Step 5 ：Solving this polynomial for x, we find the remaining zeros are -3 and 1/3.

Step 6 ：Final Answer: The remaining zeros of the function are \(\boxed{-3}\) and \(\boxed{\frac{1}{3}}\).