Write the equation of a polynomial that has 2 and 3-i as roots

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

f(x) = y

f(-3) = -5

step-by-step explanation: