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
