4   1   0   0   0 256
        4  16  64 256
---------------------
    1   4  16  64 512

This must be considered as



Change the sign of the -4 in x-4 to 4.

Start with this:

4 | 1 0  0  0 256
  |              

Bring down the 1

4 | 1 0  0  0 256
  |              
    1

Multiply the 1 that you just brought down by the 4
at the far left, get 4.  Write that 4 just below
the first 0

4 | 1 0  0  0 256
  |   4          
    1

Add the 0 and the 4, get 4, write that at the bottom
like this

4 | 1 0  0  0 256
  |   4          
    1 4

Multiply the 4 that you just wrote down at the bottom
by the 4 at the far left, get 16.  Write that 16 just below
the second 0 at the top:

4 | 1 0  0  0 256
  |   4 16       
    1 4

Add the 0 and the 16, get 16, write that at the bottom
like this:

4 | 1 0  0  0 256
  |   4 16       
    1 4 16

Multiply the 16 that you just wrote down at the bottom
by the 4 at the far left, get 64.  Write that 64 just below
the third 0 at the top:

4 | 1 0  0  0 256
  |   4 16 64    
    1 4 16

Add the 0 and the 64, get 64, write that at the bottom
like this:

4 | 1 0  0  0 256
  |   4 16 64    
    1 4 16 64

Multiply the 64 that you just wrote down at the bottom
by the 4 at the far left, get 256.  Write that 256 just below
the 256 at the top:

4 | 1 0  0  0 256
  |   4 16 64 256
    1 4 16 64

Add the 256 and the 256, get 512, write that at the bottom
like this:

4 | 1 0  0  0 256
  |   4 16 64 256
    1 4 16 64 512

Now we interpret the numbers on the bottom line
1. All but the last number are the coefficients of a
polynomial of 1 less degree than the original polynomial.

So the quotient is

   1x3 + 4x2 + 16x + 64

   and the last number in the bottom of the synthetic
   division, 512, is the remainder.  Normally we place the
   remainder over the divisor, like this:

     x3 + 4x2 + 16x + 64 + 

Edwin