You can try to use the Rational Roots theorem. If x = p/q with p and q integers that don't have any factors in common, then p will divide the constant term and q will divide the coefficent of the highest power of x.
In this case we have:
f(x) = 0
with
f(x) = x^3 - 11 x^2 + 36 x - 36
A rational zero of f thus has to be an integer that divides 36. They can thus be:
x= ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36
Since were too lazy to try out all these 18 candidates, we're going to use the following trick. Let's substitute for x some random integer r such that f(r) is some ugly number with only a few divisors. How will that help?
Consider the polynomial g(y) defined as:
g(y) = f(r+y)
We apply the Rational Roots theorem to g(y). We note that to do that you don't need to expand out f(r+y), all you need are the coefficients of y^3 and the constant term of f(1+y). The constant term is the value for y = 0, which is f(r) = ugly number with only a few divisors. The coefficient of y^3 is the same as the coefficient of x^3, so the rational roots of g(y) are those few divisors of the ugly number f(r)
Since g(y) = f(r+y), adding r to these candidates gives the possible rational roots for f.
E.g., we have that f(5) = -6. Tis means that the possible rational roots of g(y) = f(5+y) are:
y = ±1, ±2, ±3, ±6
The possible rational roots of f are thus:
x = 5+y = -1, 2, 3, 4, 6, 7, 8, 11
But the rational roots also have to be divisors of 36, therefore we can strike out any items in the list that are not divisors of 36, leaving us with:
x = -1, 2, 3, 4, 6,
If we try f(-1), we see that f(-1) = -84. If we then take r = -1 and play the same game as above, we see that the list of candidates is reduced to:
x = 2, 3, 6
These are in fact the three roots of f.
