Let #f(n) = n^3-3n^2+2n-990#
By the rational roots theorem, any rational roots of #f(n) = 0# must be of the form #p/q# where #p# and #q# are integers, #q != 0#, #p# a divisor of the constant term #-990# and #q# a divisor of the coefficient, #1#, of the term #n^3# of highest degree.
So the only possible rational roots are the factors of #990#, viz
#+-1#, #+-2#, #+-3#, #+-5#, #+-6#, #+-9#, #+-10#, #+-11#, #+-18#, #+-22#, #+-30#, #+-33#, #+-45#, #+-55#, #+-90#, #+-99#, #+-110#, #+-165#, #+-198#, #+-330#, #+-495#, #+-990#
That's rather a lot of factors to check (and I'm not sure I found all of them), so let's try to get closer:
#n^3-3n^2+2n = 990#
So try #n^3 ~= 990#, say #n ~= 10#
#f(10) = 1000-300+20-990 = -270#
#f(11) = 1331-363+22-990 = 0#
So #(n-11)# is a factor.
Divide #f(n)# by #(n-11)# to find:
#f(n) = n^3-3n^2+2n-990 = (n-11)(n^2+8n+90)#
The discriminant of #n^2+8n+90# is:
#8^2-(4*1*90) = 64 - 360 = -296#
So there are no more linear factors with real coefficients.
