A polynomial with real coefficients has 3, 2i, and -i as three of its zeros.?

If 2i is a zero, then so is -2i.

If -i is a zero, then so is i.

There are five zeros (2, ± 2i, ± i), so the least possible degree is 5.

Hello,

The result is obtained by expressing the wanted polynomial as a product of (x - root).

So if you want the roots {x₀, x₁, x₂, ... , xₐ}, you just have to do:

P(x) = (x - x₀)(x - x₁)(x - x₂) ... (x - xₐ)

which will yield a polynomial of degree a.

If any given root is a complex (i.e. of the form a+ib), you MUST also have the root a-ib to obtain a polynomial with real coefficients.

Since you are provided 2 complex roots 2i and -i. We need to add the conjugate roots -2i and i to obtain a polynomial with real coefficients.

Thus, you end with 5 roots, which would leave you with a polynomial of degree 5.

Explicatively,

Dragon.Jade :-)

as it has real coefficient

if 2i is a root then -2i should be a root

if -i is a root then i is a root

as it has 5 roots 3, 2i, -2i, i, - i

so it is having degree 5

though not required polynomial = (x-3)(x-2i)(x+2i) (x-i)(x+i) = (x-3)(x^2 + 4)(x^2 + 1)