Find an nth-degree polynomial function with real coefficients satisfying the given conditions. n = 3; 4 and 2 *i* are zeros; f(2) = -16

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  f(x) = x^3 -4x^2 +4x -16

Step-by-step explanation:

If 2i is a zero, then its conjugate, -2i, is also a zero. When p is a zero, then (x-p) is a factor. For the three given zeros, the factors are ...

  f(x) = (x -4)(x -2i)(x +2i)

  f(x) = (x -4)(x^2 +4)

  f(x) = x^3 -4x^2 +4x -16 . . . . . . multiplied out to standard form

Now, f(2) = (2 -4)(2^2 +4) = (-2)(8) = -16. This is the required value, so we're done.*

  f(x) = x^3 -4x^2 +4x -16

* If the required value were something else, then we would have to multiply the f(x) we found by some factor to make the value of f(2) come out right. That factor would be ...

  k = (desired f(2))/(f(2) for our function so far)

In this case, ...

  k = -16/-16 = 1

1. 22

2. 1

3. 7

4. 18

step-by-step explanation

e2020 got it right