Given that fx is a cubic function with zeros at 5 and 2i1 f

Given that f(x) is a cubic function with zeros at 5 and 2i+1 , find an equation for f(x) given that f(7)=3.

zeros at x=-5

x= 1 -2i

there would be another zero which is complex conjugate of 1-2i = 1 +2i

So, f(x) = k( x+5)(x -1 +2i)(x -1 -2i)

=k(x+5) (x^2 -x -2ix -x +1 +2i +2ix-2i +4)

=k(x+5)(x^2 -2x +5)

Given : f(7) = 3

3 = k(7+5)(49 -14 +5)

k = 3/(12*40) = 1/160

f(x) = (1/160)(x+5)(x^2 -2x +5)