Question

How do you find a fourth degree polynomial given roots `3i` and `sqrt6`?

Answer 1

`x^4+3x^2-54 = 0`

Explanation:

The simplest polynomial with these zeros is the quadratic:

`(x-3i)(x-sqrt(6)) = x^2-(sqrt(6)+3i)x+3sqrt(6)i`

If we want integer coefficients, then `-3i` and `-sqrt(6)` are also zeros and the simplest polynomial, which is a quartic, is:

`(x-3i)(x+3i)(x-sqrt(6))(x+sqrt(6)) = (x^2-(3i)^2)(x^2-(sqrt(6))^2)`

`color(white)((x-3i)(x+3i)(x-sqrt(6))(x+sqrt(6))) = (x^2+9)(x^2-6)`

`color(white)((x-3i)(x+3i)(x-sqrt(6))(x+sqrt(6))) = x^4+3x^2-54`

Answer 2

See below.

Explanation:

The complex conjugate root theorem states that if a polynomial has a root `a+bi`, then it also has the root `a-bi`. There are also radical conjugate roots: if `a+sqrtb` is a root, then `a-sqrtb` must also be a root.

Thus, if `3i` and `sqrt6` are roots, then `-3i` and `-sqrt6` must also be roots. Now that we have the four roots, we can establish the factors that go along with them.

`(x-3i)(x-(-3i))(x-sqrt6)(x-(-sqrt6))`

`(x-3i)(x+3i)(x-sqrt6)(x+sqrt6)`

We can now expand to find the fourth-degree polynomial. Since `(a+b)(a-b) = a^2-b^2`, we can rewrite the above expression as

`(x^2-(3i)^2)(x^2-6)`

`(x^2-9i^2)(x^2-6)`

`(x^2+9)(x^2-6)`

Expanding this, we get

`x^4-6x^2+9x^2-54`

`x^4+3x^2-54`