x^3 - 6x^2 + 7x + 2
1. Possible no. of complex roots......either none or two, at most
2. Possible number of real roots.....either one or three, at most
3. Possible rational roots = +/- 1 , +/- 2
4. 2 is a root because (2)^3 - 6(2)^2 + 7(2) + 2 = 8 - 24 + 14 + 2 = 0
The other two roots are either non-real or non-rational and can be found by synthetic division
2 [ 1 - 6 7 2 ]
2 -8 -2
-------------------------
1 -4 -1 0
So the remaining polynomial is x^2 - 4x -1
And the remaining roots are 2 + sqrt(5) and 2 - sqrt(5)
