How do you factor #n^3 - 3n^2 + 2n - 990 = 0 #?

1 Answer
Jul 24, 2015

Answer:

Use the rational roots theorem and some approximating to find:

#f(n) = n^3-3n^2+2n-990 = (n-11)(n^2+8n+90)#

Explanation:

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.