x4-x³+2x²-4x-8

The factors of 8 are 1,2,and 4, so

The possible zeros are ±1, ±2, ±4, ±8

Try 1 as a zero using synthetic division.
That is we divide by x-1

1 | 1  -1   2  -4  -8
  |     1   0   2  -2
    1   0   2  -2 -10

No, the remainder is not a zero because it is -10, not 0.

Try -1 as a zero using synthetic division.
That is, we divide by x+1

-1 | 1  -1   2  -4  -8
   |    -1   2  -4   8
     1  -2   4  -8   0

Yes, the remainder is 0, so that means x+1 is a factor and gives

    1x³-2x²+4x-8 as a quotient, so the original polynomial
factors as

   (x+1)(x³-2x²+4x-8)

So we start over this time with x³-2x²+4x-8

The factors of 8 are 1,2,and 4, so

The possible zeros are ±1, ±2, ±4, ±8

There is no need to try 1 because it was not a factor of the original
polynomial

So we try -1 as a zero using synthetic division.
That is, we divide by x+1

-1 | 1  -2   4  -8
   |    -1   3  -7  
     1  -3   7 -15

No, the remainder is not a zero because it is -15, not 0.

Try 2 as a zero using synthetic division.
That is, we divide by x-2

 2 | 1  -2   4  -8
   |     2   0   8  
     1   0   4   0

Yes, the remainder is 0, so that means x-2 is a factor and gives

    1x²+0x+4 as a quotient, so the original polynomial
factors as

   (x+1)(x-2)(x²+4)

It doesn't factor further using real numbers.

Edwin