Final answer:
To find the remaining zeros of the function f(x) = x³ − 4x² + 4x − 16 with a given zero of 2i, we use the conjugate zero -2i and divide the function by the quadratic factor x² + 4 to get the linear factor x - 4, revealing the remaining real zero at x = 4.
Step-by-step explanation:
The given zero of the cubic function f(x) = x³ − 4x² + 4x − 16 is 2i. Since the coefficients of the polynomial are real, the non-real zeros must occur in conjugate pairs. Therefore, if 2i is a zero, its conjugate, -2i, must also be a zero of the function.
To find the remaining zeros, we can use the fact that (x - 2i)(x + 2i) is a factor of f(x). Multiplying these factors gives us x² + 4, which is a quadratic factor of f(x). Dividing the original cubic function by this quadratic factor yields a linear factor, which represents the remaining real zero of the polynomial.
Let's divide f(x) by x² + 4 to get the linear factor:
f(x) / (x² + 4) = x - 4
So our remaining zero is at x = 4. Now we have all the zeros of the polynomial: 2i, -2i, and 4.