x4+2x3+x2+18

We use synthetic division:

1-i√2|1    2      1       0       18
     |     1-i√2  1-4i√2 -6-6i√2 -18
      1    3-i√2  2-4i√2 -6-6i√2   0

So we have now factored the polynomial as

[x-(1-i√2)][x³+(3-i√2)x²+(2-4i√2)x+(-6-6i√2)]

Now we know that since 1-i√2 is a zero, so is
its conjugate 1+i√2, so we divide the third 
degree polynomial synthetically, this time by 1+i√2

1+i√2|1    3-i√2  2-4i√2 -6-6i√2
     |     1+i√2  4+4i√2  6+6i√2
      1    4      6       0

So now we have factored the polynomial as

[x-(1-i√2)][x-(1+i√2)](x²+4x+6)

Now we find the zeros of x²+4x+6 by the quadratic formula

x²+4x+6 = 0

x = 

x = 

x = 

x = 

x = 

x = 

x = 

x = 

x = -2±i√2

So -2+i√2 and -2-i√2 are solutions.

Thus we have now factored the entire polynomial as

[x-(1-i√2)][x-(1+i√2)][x-(2-i√2)][x-(-2+i√2)]

and the four zeros are all imaginary.  They are:

1-i√2, 1+i√2, 2-i√2, -2+i√2

Edwin