Final answer:
To find f(4), for the third degree polynomial f(x) with zeros at 3 and 5i, we express it as (x - 3)(x - 5i)(x + 5i), simplify, and substitute x with 4. The resulting value of the function at x=4 is 41.
Step-by-step explanation:
A third degree polynomial function can be written in the form f(x) = a(x - r)(x - s)(x - t), where r, s, and t are the zeros of the polynomial. In this case, the zeros are x = 3 and x = 5i. Since 5i is a complex number, its conjugate -5i must also be a zero. So we have r = 3, s = 5i, and t = -5i.
Since the leading coefficient is 1, we know that a = 1. Therefore, the polynomial is f(x) = (x - 3)(x - 5i)(x + 5i).
To find f(4), we substitute x = 4 into the polynomial: f(4) = (4 - 3)(4 - 5i)(4 + 5i).
Simplifying, we get f(4) = (1)(-1)(9) = -9.
The question asks to find the value of a third degree polynomial function f(x) at x=4, given that the function has zeros at x=3 and x=5i, and a leading coefficient of 1. Since the function is of third degree and has complex roots, we must also consider the complex conjugate of 5i, which is -5i, as a zero of the function. Therefore, the polynomial can be expressed as f(x) = (x - 3)(x - 5i)(x + 5i). Simplifying, we obtain f(x) = (x - 3)(x2 + 25).
To find f(4), we substitute x with 4: f(4) = (4 - 3)(42 + 25). This simplifies to f(4) = (1)(16 + 25) = 41. Therefore, the value of the function at x=4 is 41.