For this specific polynomial (call it p(x)), it's easy to see that x=1 is a root (p(1)=0), since the sum of the coefficients of p(x) is zero (1+1-2=0). Therefore x-1 is a factor of p(x). And using either long or synthetic division, we get the other factor (x^2+x+2), whose roots can be found using the quadratic formula.
