Question

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

Answer 1

The theorem states that the polynomial has no rational roots.

Explanation:

The rational root theorem states that if you have a polynomial with integer coefficients (as you have), then a solution `x=p/q`, where `p` and `q` have no common divisors, must be such that `p` divides `a_0`, and `q` divides the leading coefficient.

Since in your case the leading coefficient is the coefficient of `x^3`, which is one, then `q` must divide one, and so it must be one itself. On the other hand, `p` must divide `-3`, which means that it can be `-3,-1,1,3`.

So, we can check these four values, because we know what if there's a solution, it must be one of these four.

Let `p(x)=x^3-x^2-x-3`. Then,

• `p(-3)=(-3)^3 - (-3^2) - (-3) -3 =`

`-27 - 9 - (-3) -3 =`

`-27-9+3-3 = -36 ne 0`

• `p(-1)=(-1)^3 - (-1^2) - (-1) -3 = -1-1+1-3 = -4 ne 0`
• `p(1)=1^3 -1^2 - 1 -3 = 1-1-1-3 = -4 ne 0`
• `p(3)=3^3 -3^2 - 3 -3 = 27-9-3-3 = 12 ne 0`

This means that the polynomial has no rational roots.