If R is a zero then (x-R) is a factor
If A+Bi is a zero, and the coefficients are real numbers, then A-Bi is
also a solution.
There is always a constant k which is a factor (it may be 1 understood but
it may not be)

f(x) = k(x-2)(x-2i)(x+2i)
f(x) = k(x-2)(x²-4i²)
f(x) = k(x-2)(x²-4∙-1)
f(x) = k(x-2)(x²+4)
f(x) = k(x³+4x-2x²-8)
f(x) = k(x³-2x²+4x-8)
f(-1) = k[(-1)³-2(-1)²+4(-1)-8] = -15
                 k[-1-2(1)-4-8] = -15
                    k[-1-2-4-8] = -15
                         k[-15] = -15
                              k = 1
f(x) = 1(x³-2x²+4x-8)
f(x) = x³-2x²+4x-8        <--final answer

Here's the graph.  Scroll down to see it all.




Edwin