Finding an nth-degree polynomial function with real coefficients .

Question

Given information :

n=3

2 and 4 i are zeros;

f(1)=-51

Answer 1

2 and 4i are zeros

-4i is another zero, as imaginary zeros always come in conjugate pairs

x=+/-4i

x^2 = -16

x^2+16 is one factor

x-2 is another factor

a(x-2)(x^2+16) =f(x)

f(1) =a(1-2)(1+16) = -17a=-51

a = 51/17= 3

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

= 3(x^3-2x^2+16x -32)

=3x^3 -6x^2 +48x -96

is the 3 degree (n=3) polynomial with zeros -3 and -4i, where f(1)=-51

f(1) = 3(1)^3-6(1)^2+48(1)-96= -51