4x³-10x²-8x+6
Factor out 2
2(2x³-5x²-4x+3)
All the candidates for rational zeros are
±1, ±1/2, ±3, ±3/2
We try 1 with synthetic division:
1 | 2 -5 -4 3
| 2 -3 -7
2 -3 -7 -4
That doesn't work because the last number on
the bottom is -4 not 0.
We try -1
-1 | 2 -5 -4 3
| -2 7 -3
2 -7 3 0
So x+1 is a factor. so
2(2x³-5x²-4x+3) becomes
2(x+1)(2x²-7x+3)
The quadratic factors and the complete factorization is:
2(x+1)(x-3)(2x-1)
Edwin
