+140 i need to completely factor 2x4-x3-7x-10
i figured out the roots were -1 and 2
i divided the polynomial by x-2 and got (x-2)(2x3+3x2+6x+5)=2x4-x3-7x-10
then i divided that by x+1 and got (x-2)(x+1)(2x2+x+5)=2x4-x3-7x-10
and am stuck
Factor the following:
2 x^4-x^3-7 x-10
The possible rational roots of 2 x^4-x^3-7 x-10 are x = ±1/2, x = ±5/2, x = ±1, x = ±2, x = ±5, x = ±10. Of these, x = -1 and x = 2 are roots. This gives x+1 and x-2 as all linear factors:
((x+1) (x-2) (2 x^4-x^3-7 x-10))/((x+1) (x-2))
| |
x | - | 2 | | 2 x^3 | + | 3 x^2 | + | 6 x | + | 5
2 x^4 | - | x^3 | + | 0 x^2 | - | 7 x | - | 10
2 x^4 | - | 4 x^3 | | | | | |
| | 3 x^3 | + | 0 x^2 | | | |
| | 3 x^3 | - | 6 x^2 | | | |
| | | | 6 x^2 | - | 7 x | |
| | | | 6 x^2 | - | 12 x | |
| | | | | | 5 x | - | 10
| | | | | | 5 x | - | 10
| | | | | | | | 0:
(2 x^3+3 x^2+6 x+5)/(x+1) (x+1) (x-2)
| |
x | + | 1 | | 2 x^2 | + | x | + | 5
2 x^3 | + | 3 x^2 | + | 6 x | + | 5
2 x^3 | + | 2 x^2 | | | |
| | x^2 | + | 6 x | |
| | x^2 | + | x | |
| | | | 5 x | + | 5
| | | | 5 x | + | 5
| | | | | | 0:
Answer: |2 x^2+x+5 (x+1) (x-2)
