Nth degree polynomial function

Question

n = 3 ; 4 and 4 i are zeros f (1) = - 153

Answer 1

x=3, x-3 is one factor

x=4, x-4 is a 2nd factor

x = 4i

x^2 = -16

x^2+16 is a 3rd factor with 2 conjugate zeros 4i and -4i

multiply all 3 factors together

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

(x^2 -7x +12)(x^2 +16)

= x^4- 6x^3 + 28x^2 - 112x + 192

f(1)=-153 = a(1 -6 +28 -112 +192) = a(103(

a = -103/153

f(x) = (-103/153)(x^4 -6x^3 + 28x^2 -112x + 192