Question

How do you use the rational root theorem to find the roots of `x^3 – x^2 – x – 3 = 0`?

Answer 1

The rational root theorem only helps us find that `x^3-x^2-x-3 = 0` has no rational roots.

The only Real root is:

`x = 1/3+root(3)(46+6sqrt(57))/3+root(3)(46-6sqrt(57))/3 ~~ 2.1304`

Explanation:

Let `f(x) = x^3-x^2-x-3`.

By the rational root theorem, any rational roots of `f(x) = 0` must be of the form `p/q` in lowest terms, where `p, q in ZZ`, `q != 0`, `p` is a divisor of the constant term `-3` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational roots of `f(x) = 0` are:
`+-1`, `+-3`

Let's try them:

`f(1) = 1 - 1 - 1 -3 = -4`
`f(-1) = -1-1+1-3 = -4`
`f(3) = 27-9-3-3 = 12`
`f(-3) = -27-9+3-3 = -36`

So `f(x) = 0` has no rational roots and the rational root theorem cannot help us find the roots that do exist.

Using a simple Tschirnhaus transformation (`t = x - 1/3`) and Cardano's method, I found:

`x = 1/3+root(3)(46+6sqrt(57))/3+root(3)(46-6sqrt(57))/3 ~~ 2.1304`