p(x) = x^3 - 6x^2 + 11x -13
= x^3 -6x^2 + 12x - 8 - x - 5
= (x-2)^3 - x + 2 - 7
= (x-2)^3 - (x-2) - 7
= X^3 - X - 7, where X = x-2
you can go from 1st to second line by letting x^3 - 6x^2 be the first 2 terms of a binomial raised to the third power. ie. general form X^3 + 3a x^2 + 3a^2 x + a^3. This lets you know how to split up the other terms.
let q(X) = X^3 - X - 7 [ = X(X^2 - 1) -7]
q'(X) = 3X^2 - 1
solutions of q'(X) = 0 are X = +/- 1/sqrt3
q(1/sqrt3)*q(-1/sqrt3) = {1/sqrt3 (1/3 - 1) -7 }{-1/sqrt3 ( 1/3-1) - 7}
={1/sqrt3 (-2/3) - 7}{-1/swrt3 (-2/3) - 7}
= (-7)^2 - (2/3sqrt3)^2
= 49 - 4/27 > 0
therefore q(1/sqrt3) and q(-1/sqrt3) are either both positive or both negative. therefore both stationary points exist on the same side of the x-axis. therefore there is only one real root (examine a graph of a cubic and see that if both stationary points are on the same side, there can only be one point where the graph crosses the x-axis).
in general if u got a cubic in the form p(x) = x^3 - ax + b => p'(x) = 3x^2 - a
solutions of p'(x) = 0 are x = +/- sqrt(a/3). Let them be A and B.
then consider the term p(A)p(B).
if p(A)p(B) > 0 then both stationary points have the same sign thus 1 real solution.
if p(A)p(B) < 0 then the stationary points are on opposite sides of the x-axis, thus 3 real roots.
if p(A)p(B) = 0 then there is a real double root, and another real root.
note also if a<0 then p'(x) = 0 has no solutions and the cubic has no stationary points, thus one real root.
and if a=0, p(x) is just x^3 + b and that's quite easy to analyse.
