Question

State how many imaginary and real zeros the function has.

Answer 1

The answer is a)
0 imaginary; 3 real
proof

f(3)= = 27-20.9+123.3-216= 216-216=0
so f(x)=(x-3)(ax²+bx+c)=0
after doing euclidean division
ax²+bx+c= (x-8)(x-9)=x²-17x+72
and  x²-17x+72 = (x-8)(x-9)
so f(x)=(x-3)(x-8)(x-9)

f(8)= 8^3-20*64+123*8-216=216-216=0
f(9)=9^3-20*81+123*9-216=0