Answer:
Step-by-step explanation:
complex roots always occur in pairs.
roots are 2,3i,-3i
f(x)=(x-2)(x-3i)(x+3i)
=(x-2)(x²-(3i)²)
=(x-2)(x²-9i²)
=(x-2)(x²+9)
=x³-2x²+9x-18
if the roots are 2±3i or 2+3i and 2-3i
f(x)=(x-(2+3i))(x-(2-3i))
=[(x-2)-3i][(x-2)+3i]
=[(x-2)²-(3i)²]
=x²-4x+4-9i²
=x²-4x+4+9
=x²-4x+13
