The critical piece of knowledge that enables one to do this problem is that complex roots ALWAYS appear as conjugate pairs. So if 1 - 2i is a zero, then 1 + 2i is also a zero. We then have all three of the zeros for the cubic function (by the Fundamental Theorem of Algebra, each 3rd degree polynomial has in total 3 zeros).
To find the polynomial with zeros b, c, and d we just multiply together the associated binomials so:
P(x) = a(x - b)(x - c)(x - d). Note that the simplest version of this polynomial (in my opinion) is when a = 1. That makes the polynomial for this case:
P(x) = (x - 1)[x - (1 - 2i)][x - (1 + 2i)]
P(x) = (x - 1)(x - 1 + 2i)(x - 1 - 2i)
To multiply these, a generic rectangle is useful. I'll first multiply the two complex parts:
Notice that we get terms that zero out. 2ix cancels with -2ix and 2i cancels with -2i. We can also simplify -4i2 since i2 = -1, then -4i2 = -4(-1) = 4 and we combine like terms to get x2 - 2x + 5.
Then multiply (x2 - 2x + 5) by (x - 1) in the same way to get: P(x) = x3 - 3x2 + 7x - 5