Let's use synthetic division.  [Yes, you can use synthetic division
even with complex numbers!  However we'll have to stop along the way
to multiply two complex numbers.]


4-i | 1   -8   19   -16   34
    |      4                 
      1   -4-i

We stop and multiply -4-i by 4-i: (-4-i)(4-i) = -16+4i-4i+i² = -16+(-1) = -17

4-i | 1   -8    19  -16    34
    |      4-1 -17    8-2i   
      1   -4-i   2   -8-2i

We stop and multiply -8-2i by 4-i: (-8-2i)(4-i) = -32+8i-8i+2i² = -32+2(-1) = -34

4-i | 1   -8    19  -16     34
    |      4-1 -17    8-2i -34
      1   -4-i   2   -8-2i   0

We have factored the equation as

[x-(4-i)][x³ + (-4-i)x² + 2x +(-8-2i)] = 0

Now since 4-i is a root, so is its conjugate 4+i. Now we divide
the bracketed third degree polynomial by [x-(4+i)]

4+i | 1   -4-i   2    -8-2i
    |      4+i   0     8+2i
      1      0   2        0

Now we have further factored the polynomial equation as


[x-(4-i)][x-(4+i)](x²+2)

We can factor x²+2 by writing it as x²-(-2) = x²-(√-2)² = (x-√-2)(x+√-2) = 
(x-i√2)(x+i√2)}}}

So the final factorization is:

[x-(4-i)][x-(4+i)](x-i√2)(x+i√2) = 0 

So the 4 solutions are 4-i, 4+i, i√2, -i√2

Edwin