There's nothing special about the (x - 3 + 2i) factor, for instance. All it's saying is that I want a term that equals zero when "3 - 2i" is substituted for x.
The other imaginary term pops in there because the complex conjugate will always be a root.
The whole point is that is from the product (x-3+2i)(x-3- 2i)(x-3)(x-3). For this problem, at least, it is not a matter of "factoring" but of "multiplying"!
As far as (x-(3- 2i))(x-(3+2i))= (x-3+2i)(x-3-2i) is concerned, think of it as a "product of sum and difference= difference of squares":