Find a polynomial function in standard form that has zeros -2, 3, 2i

Question

anonymous · Answer

If it has a zero at those points, then we can say that in factored form, it has a term that is zero at that point.

That is, if a, b, and c are zeroes, we can say a factored form of it is:
(x-a)(x-b)(x-c) = 0

Then we can distribute this out into standard form.

anonymous · Answer

so a being -2, b being 3, and c being 2i?

anonymous · Answer

Ignore the "= 0" part. Not sure why I put it there.

But yeah, just plug those in.

anonymous · Answer

The 2i is throwing me off

anonymous · Answer

It's just a number. Just remember that you can't add/subtract real numbers and imaginary numbers, but you can multiply/divide them.

anonymous · Answer

When simplifying, that is.

hero · Answer

That's because there's another zero that's the conjugate of 2i

anonymous · Answer

Is the -2i?

anonymous · Answer

So would that be (x+2)(x-3)(x-2i)?

hero · Answer

"There's ANOTHER zero" (in addition to 2i) which is the conjugate of 2i. In other words, there's another factor to consider:

(x + 2)(x - 3)(x + 2i)(x - 2i) = 0

anonymous · Answer

That's what I was missing! I knew the i's had to cancel somewhere....Thank you!

hero · Answer

Happy to help bro. Good luck with your work.