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 states that any rational root of a polynomial will be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.
In our case `q=1` and `p=-3` hence possible roots are

`1,-1,-3,3`

But if you check none of the values above are roots for
`x^3-x^2-x-3=0`.
So the rational root theorem cannot help us here.

Answer 2

You can't. You can only use the rational root theorem to show that it has no rational roots.

Explanation:

By the rational root theorem, any rational roots of `x^3-x^2-x-3=0` must be expressible in the form `p/q` for integers `p` and `q` with no common factor apart from `+-1`, where `p` is a factor of the constant term `-3` and `q` a factor of the coefficient `1` of the leading term.

That means that the only possible rational factors are:

`+-1`, `+-3`

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

We find:

`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 `x^3-x^2-x-3=0` has no rational roots.

In fact it has one Real root:

`x = 1/3 (1+(46-6 sqrt(57))^(1/3)+(46+6 sqrt(57))^(1/3))`

and two Complex roots:

`x = 1/3 (1+omega(46-6 sqrt(57))^(1/3)+omega^2(46+6 sqrt(57))^(1/3))`

`x = 1/3 (1+omega^2(46-6 sqrt(57))^(1/3)+omega(46+6 sqrt(57))^(1/3))`

where `omega = -1/2+sqrt(3)/2 i` is the primitive Complex cube root of `1`.

These can be found using Cardano's method or similar.