Let #f(x) = x^3-2x^2-5x+6#

First note that the sum of the coefficients is #0#, so #x=1# is a zero of #f(x)#...

#f(1) = 1-2-5+6 = 0#

Divide #f(x)# by #(x-1)# to find:

#f(x) = x^3-2x^2-5x+6 = (x-1)(x^2-x-6)#

Then #x^2-x-6 = (x-3)(x+2)#

So the other two roots of #f(x) = 0# are #x=3# and #x=-2#

Alternatively, use the rational roots theorem to know that any rational roots of #f(x) = 0# must be of the form #p/q# where #p#, #q# are integers, #q != 0#, #p# a factor of the constant term #6# and #q# a factor of the coefficient #1# of the term (#x^3#) of highest degree. So the only possible rational roots are:

#+-1#, #+-2#, #+-3# and #+-6#

Try each to find #f(1) = 0#, #f(-2) = 0# and #f(3) = 0#.