We use the fact that if a polynomial function has an imaginary
zero, such as 1+i, then its conjugate, 1-i, is also a zero
of the polynomial.  That gives us three different zeros, and so
it will be of degree 3.  We begin by setting x = to each of the
three zeros:

  x = 1+i,     x = i-i,  x = -4

We get 0 on the right of each:

x-1-i = 0, x-1+i = 0,  x+4 = 0

Now we multiply the three left sides and multiply the three right 
sides, which just gives us 0 on the right:

(x-1-i)(x-1+i)(x+4) = 0

We do some grouping on the first two:

[(x-1)-i][(x-1)+i](x+4) = 0

We multiply the first two bracket expressions using 
the principle (A-B)(A+B) = A²-B²: 

       [(x-1)²-i²](x+4) = 0

We use the fact that i² = -1:

       [(x-1)²-(-1)](x+4) = 0

       [(x-1)²+1](x+4) = 0

We FOIL that out:

  (x-1)²x+4(x-1)²+1x+4 = 0 

We use the fact that (x-1)² = (x-1)(x-1) = x²-x-x+1 = x²-2x+1

  (x²-2x+1)x+4(x²-2x+1)+x+4 = 0

      x³-2x²+x+4x²-8x+4+x+4 = 0

                x³+2x²-6x+8 = 0

Since that polylomial equation has solutions (roots) 1+i, 1-i, and -4
The polynomial function

             f(x) = x³+2x²-6x+8

has those three zeros.

Edwin