Working with $i$ and finding roots How do you find the cubic polynomial $f(x)$ with integer coefficients, where $1-3i$ and $-2$ are the roots of $f(x)$?
It has been too long since I have worked with $i$ and I have been unable to generate it.

The other solutions correctly point out that the third root must be $1+3i$, and therefore the polynomial must be $$(x-(1-3i))(x-(1+3i))(x-2)$$
or a multiple of it (notice that multiplying through the entire thing by a constant does not change the zeroes).
I will add to this one tip for simplifying the expression:  Notice that you can regroup and multiply out the factors that involve $i$ as follows:
$$(x-(1-3i))(x-(1+3i)) = ((x-1) + 3i)((x-1) - 3i) = (x-1)^2 + 9$$
which might make the problem slightly easier to complete.
