hello :
if : -2i the zero of f(x) then : 2i is zero ...( even function )
so : f(x) = (x-2i)(x+2i) g(x)..... g(x) = ax²+bx +c
butt : (x-2i)(x+2i) = x²-(2i)² =x²+4... (i²=-1)
f(x) = (x²+4)(ax²+bx+c) = ax^4 +bx^3+cx² +4ax² +4bx+4c
f(x) = ax^4 +bx^3 +(c+4a)x²+4bx+4c
but : f(x) = x4 - 5x2 - 36
so :
a =1
b=0
c+4a=-5
4c = -36
conclusion : a= 1 b= 0 c= -9
g(x) x²-9
g(x) = 0 : x²-9 =0
(x-3)(x+3) = 0
x-3=0 or x+3=0
x=3 or x= -3
all zero are : 2i , -2i , 3 , -3
