The number of rational roots of polynomial $ x^3 - 3x - 1$ is?

Question

How exactly can I find no. of rational roots, not only for this but similar type of questions?

Answer 1

We will prove that this equation has no rational roots by contradiction.

Suppose $x= \dfrac pq$ (Rational) is a root of this equation , where $\gcd (p,q) =1$

Now put the value of $x$ in the equation :

$$ \Big (\frac{p}{q} \Big )^3 - 3 \Big (\frac{p}{q} \Big ) -1=0 \implies p^3-3pq^2-q^3=0$$

Now take a look at the following cases :

Case 1 : If $p$ is even $\implies$ $q$ is also even. Not possible , since $\gcd (p,q) =1$

Case 2 : If $p$ is odd, and $q$ is also odd , This expression can never be zero. (odd $-$ odd $-$ odd $\neq 0$)

Case 3 : If $p$ is odd, and $q$ is even, again, this expression can never be zero. (odd $-$ even $-$ even $\neq 0$)

Our assumption was wrong.

Therefore, this equation has no rational root.

Answer 2

There is a theorem which states that all and only rational roots of a polynomial with integer coefficients are to be searched among the ones of the form $\frac ab$ where $a\in\Bbb Z$ divides the known term and $b\in\Bbb Z$ (obviously $b\neq0$) divides the coefficient of the highest power.

So in your case all and only rational roots could be $\pm1$, but plugging them into your polynomial, you'll easily see that none of them is a root.