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

Nov 10, 2015

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 $x = \frac{p}{q}$, 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 $p \left(x\right) = {x}^{3} - {x}^{2} - x - 3$. Then,

• $p \left(- 3\right) = {\left(- 3\right)}^{3} - \left(- {3}^{2}\right) - \left(- 3\right) - 3 =$

$- 27 - 9 - \left(- 3\right) - 3 =$

$- 27 - 9 + 3 - 3 = - 36 \ne 0$

• $p \left(- 1\right) = {\left(- 1\right)}^{3} - \left(- {1}^{2}\right) - \left(- 1\right) - 3 = - 1 - 1 + 1 - 3 = - 4 \ne 0$
• $p \left(1\right) = {1}^{3} - {1}^{2} - 1 - 3 = 1 - 1 - 1 - 3 = - 4 \ne 0$
• $p \left(3\right) = {3}^{3} - {3}^{2} - 3 - 3 = 27 - 9 - 3 - 3 = 12 \ne 0$

This means that the polynomial has no rational roots.