x-2; 4x³-3x²-8x+4
Change the sign of the number -2 in x-2, to +2.
Put it on the left of the synthetic division:
2 | 4 -3 -8 4
| 8 10 4
4 5 2 8
The last number on the bottom row, which is the remainder,
is not 0, it's 8. Therefore x-2 is not a factor of
4x³-3x²-8x+4.
---------------------
Note:
If the polynomial had been 4x³-3x²-8x-4 (where the last term
were -4 instead of +4, then the synthetic division would have
been:
2 | 4 -3 -8 -4
| 8 10 4
4 5 2 0
Then x-2 would have been a factor of 4x³-3x²-8x-4, because
the last term, which is the remainder, would have been 0
and the factorization would have been
(x-2)(4x²+5x+2) using the numbers on the bottom row as
coefficient of a polynomial of one less degree.
However with the +4 on the end x-2 is not a factor.
Edwin
