Use the given zero to find the remaining zeros of the function. \[ h(x)=7 x^{5}+3 x^{4}+140 x^{3}+60 x^{2}-875 x-375 \text { zero: }-5 i \]

Step 1 ：Given that -5i is a zero of the polynomial, we know that its complex conjugate, 5i, is also a zero due to the Conjugate Root Theorem. This theorem states that if a polynomial has real coefficients and a complex number is a root of that polynomial, then its conjugate is also a root.

Step 2 ：We can then divide the polynomial by the quadratic factor corresponding to these two zeros to find the remaining zeros. The quadratic factor corresponding to the zeros -5i and 5i is \(x^2 + 25\), because \((x - (-5i))(x - 5i) = x^2 + 25\).

Step 3 ：The result of the division is a cubic polynomial and a remainder of 0, which means the division is exact and the quadratic factor \(x^2 + 25\) is indeed a factor of the original polynomial. The cubic polynomial \(7x^3 + 3x^2 - 35x - 15\) represents the remaining part of the polynomial after accounting for the zeros -5i and 5i.

Step 4 ：We can find the remaining zeros by setting this cubic polynomial equal to zero and solving for x.

Step 5 ：The remaining zeros of the function are \(-\frac{3}{7}\), \(-\sqrt{5}\), and \(\sqrt{5}\).

Step 6 ：\(\boxed{\text{Final Answer: The complete set of zeros of the function is } -5i, 5i, -\frac{3}{7}, -\sqrt{5}, \text{ and } \sqrt{5}.}\)