In #x^3+x^2-4x+6#, zeros of the function would be factors of #6# i.e. among #{1,-1,2,-2,3,-3,6,-6}#.

It is apparent that one such zero is #-3# as putting #x=-3# in #x^3+x^2-4x+6#, we get #(-3)^3+(-3)^2-4(-3)+6=-27+9+12+6=0#.

Hence #(x+3)# is a factor of #x^3+x^2-4x+6# and dividing latter by former we get #x^2-2x+2#, for which discriminant is #(-2)^2-4*1*2=-4# and hence cannot be factorized with real coefficients.

Hence, the only zero of #x^3+x^2-4x+6# is #-3#.

However, if domain of #x# is extended to complex numbers, we will also have #(2+-sqrt(-4))/2# (as they are zeros of #x^2-2x+2# using quadratic formula) or #1+2i# and #1-2i#.