Question

Fine a third degree polynomial function f(x) with real coefficients that has 4 and 2i are zeros and such that f(-1) =-50
4 21 -2i
(x-4)(x-2i)(x+2i)
(x-4)(x2+4)
x3-16+4x-4x2
(x3-4x2+4x-16)
-50=a(-1-4-4-16)
-50=.25
a=2
Not understanding, please help

Answer 1

I assume you get this far ok:
...
(x-4)(x^2+4)
x^3-4x^2+4x-16
evaluate that at x = -1, and you get
-1-4-4-16 = -25
That means that if
f(x) = a(x^3-4x^2+4x-16)
then f(-1) = -25a

But, we want f(-1) = -50
so, -25a = -50
a = 2

f(x) = 2(x^3-4x^2+4x-16)