Final answer:
An nth-degree polynomial function with 4i as a zero is a fourth-degree polynomial, given that complex zeros occur in conjugate pairs. The polynomial can be expressed as (x - 4i)(x + 4i) squared. Without additional information, we cannot satisfy the condition f(-1) = 102.
Step-by-step explanation:
To find an nth-degree polynomial function with real coefficients that has 4i as a zero, we need to first recognize that complex zeros in polynomials with real coefficients come in conjugate pairs. This means that if 4i is a zero, then its conjugate, -4i, is also a zero. Since a polynomial of degree n has exactly n zeros (counting multiplicity), we are looking for a fourth-degree polynomial, which implies we have two more zeros to find. However, the question does not provide additional zeros, so we will have to express our polynomial with the two zeros we know.
The polynomial function can be described by the product of its factors. Starting with the zeros 4i and -4i, we get two factors: (x - 4i) and (x + 4i). Multiplying these factors will give us a quadratic equation with real coefficients: (x - 4i)(x + 4i) = x2 + 16. Because we need a fourth-degree polynomial, we can square this result to get our polynomial: f(x) = (x2 + 16)2.
Regarding the condition f(^-1) = 102, it's unclear what is meant by f(^-1) due to the notation used; it might be a typo. Assuming it refers to f(-1), which is a common way to specify the value of the function when x = -1, we would substitute x = -1 into our polynomial to see if it equals 102. Doing so gives us f(-1) = ((-1)2 + 16)2 = 172 = 289, which does not equal 102. Without additional zeros or modifications, we cannot satisfy the condition f(-1) = 102 with the given information.