Question

How do you find a third degree polynomial given roots `3` and `2-i`?

Answer 1

We are given roots

`x_1=3`
`x_2=2-i`

The complex conjugate root theorem states that, if `P` is a polynomial in one variable and `z=a+bi` is a root of the polynomial, then `bar z=a-bi`, the conjugate of `z`, is also a root of `P`.

As such, the roots are

`x_1=3`
`x_2=2-i`
`x_3=2-(-i)=2+i`

From Vieta's formulas, we know that the polynomial `P` can be written as:

`P_a(x)=a(x-x_1)(x-x_2)(x-x_3)`

Where `a` is a constant.

`:. P_a(x) = a(x-3)(x-2+i)(x-2-i)`

`:. P_a(x) =a(x^3-7x^2+17x-15)`

One such polynomial would be

`P_1(x) = x^3-7x^2+17x-15`