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

1 Answer
George C.
Jul 24, 2015

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.

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