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

9514 1404 393

  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

i dont know if you cant find a way its best to actually add all of them start from 105 tho

step-by-step explanation:

a

step-by-step explanation: