Question

what is the polynomial with roots 2, -3i, 3i

Answer 1

(x-2)(x+3i)(x-3i)=0

Answer 2

It has to be multiplied out and that's the part I don't know how to do

Answer 3

(x+3i)(x-3i ) = x^2 + 9

Answer 4

Let (x-2)(x+3i)=a. Use the distributive property:
a(b+c)=ab+bc
a(x-3i)=ax-a3i=xa-3ia
Now let's multiply a out. Let (x-2)=p
a=p(x+3i)=xp+3ip=x(x-2)+3i(x-2)=x^2-2x+3i(x-2)=x^2-2x+3ix-6i=x^2+(3i-2)x-6i
Substitute that above
(xi2)(x+3i)(x-3i)=x(x^2+(3i-2)x-6i)-3i(x^2+(3i-2)x-6i)=x^3+(3i-2)x^2-6ix-3ix^2+(-9+6i)x+18=\(x^3-2x^2-9x+18\)