Write an equation of lowest degree with real coefficients with the given zeros

root2, 3i
Please help! I don't understand this at all..

if the coefficients are real, then complex roots come in conjugate pairs. So, if 3i is a root, so is -3i

So the polynomial is (x-√2)(x-3i)(x+3i) = (x-√2)(x^2+9)

If the coefficients are rational, then surds also come in conjugate pairs, so if √2 is a root, so is -√2

That means our polynomial is
(x-√2)(x+√2)(x-3i)(x+3i) = (x^2-2)(x^2+9)

To find an equation of lowest degree with real coefficients that have the given zeros, we can use the fact that complex zeros always occur in conjugate pairs.

First, we have the real zero 'root2'. Therefore, one factor of the equation can be (x - root2).

Now, we have the complex zero '3i'. Since complex zeros occur in conjugate pairs, the conjugate of '3i' is '-3i'. So, the factors for the complex zeros are (x - 3i) and (x + 3i).

To get the equation of lowest degree, we multiply all the factors together:

(x - root2) * (x - 3i) * (x + 3i)

Multiplying these factors out gives us:

(x - root2)(x^2 + 9)

Expanding further:

(x^3 - root2x^2 + 9x - root2 * 9)

Hence, an equation of lowest degree with real coefficients and the given zeros is:

x^3 - root2x^2 + 9x - root2 * 9