Final answer:
The third degree polynomial function with zeros at x=-2 and x=i and leading coefficient 1 is f(x)=(x+2)(x^2+1). To find f(3), we evaluate it at x=3 resulting in f(3)=50.
Step-by-step explanation:
The given question requires finding the value of the third degree polynomial function f(x) at x=3, knowing that it has zeros at x=-2 and x=i, and its leading coefficient is 1. Since the polynomial has a complex zero x=i, its conjugate x=-i must also be a zero due to the Complex Conjugate Root Theorem. This gives us the three factors of f(x): (x+2), (x-i), and (x+i). The function can thus be written as:
A polynomial function of degree three 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 and a is the leading coefficient. Since the polynomial has zeros at x = -2 and x = i, we can write the function as f(x) = (x + 2)(x - i)(x + i). To find f(3), we substitute x = 3 into the function: f(3) = (3 + 2)(3 - i)(3 + i). Simplifying further gives us the value of f(3).f(x)=(x+2)(x-i)(x+iThe function simplifies to:(x)=(x+2)(x^2+1)To find f(3), we substitute x with 3f(3)=(3+2)((3)^2+1)=(5)(9+1)=(5)(10)=50Thus, f(3) is 50.