find 3rd degree with real coefficient and with zeros at 1 and 2i?

Question

find the 3rd degree with real coefficient and with zeros at 1 and 2i?

I have no clue how to do this. It's on my test but I don't think we've learned it. Please explain.

Answer 1

Imaginary roots (zeroes) comes in pairs so -2i is also a root.
Multiply:
(x-1)(x+2i)(x-2i)=
(x-1)(x^2+4)=
x^3-x^2+4x-4

Answer: x^3-x^2+4x-4

Answer 2

Write a polynomial function of least degree that has real coefficients, the given zeros, and a leading coefficient of 1. 1. 5, 2i, -2i (x-5)*(x+2i)*(x-2i) multiply it out

Answer 3

If 2i is a zero then so must be -2i making two of the factors be (x - 2i) and (x + 2i) which multiply out to be x^2 - 4i^2 which is just (x^2 + 4). And if 1 is a zero then (x-1) is a factor. So multiply (x^2 + 4)(x - 1) and you'll get it.

Answer 4

With an imaginery root, you must have it's opposite as well:
+- 2i
(x - 2i)(x + 2i)(x - 1)
(x^2 + 4)(x - 1)
x^3 - x^2 + 4x - 4

Answer 5

If the polynomial, let's call it P(x), has real
coefficients and a+ bi is a root, so is a-bi.
So your roots are 1, 2i and -2i.
So P(x) = (x-1)(x+2i)(x-2i) = (x-1)(x²+4) = x³-x²+4x-4.