By the rational roots theorem, any rational roots of #f(x) = 0# must be of the form #p/q# where #p#, #q# are integers, #q != 0#, #p# a divisor of the constant term #22# and #q# a divisor of the coefficient #1# of the term of highest degree #x^3#.

So the only possible rational roots are:

#+-1#, #+-2#, #+-11#, #+-22#

#f(1) = 1+10-13-22 = -24#

#f(-1) = -1+10+13-22 = 0#

#f(2) = 8+40-26-22 = 0#

So far, that gives #(x+1)# and #(x-2)# as factors of #f(x)#.

So the remaining factor must be #(x+11)# to get the correct coefficient for #x^3# and the constant term #-22#.

#f(x) = (x+1)(x-2)(x+11)#

Let us check:

#f(-11) = -1331+1210-143-22 = 0#

So the roots of #f(x) = 0# are #x = -1#, #x=2# and #x=-11#