# $P(x)$ doesn't have a rational root?

Let $$P(x)$$ be a polynomial with integer coefficients. In what conditions that $$P(x)$$ doesn't have a rational root?

From https://en.wikipedia.org/wiki/Rational_root_theorem, if $$P(x)=a_nx^n+\cdots +a_0$$ with integer coefficients, where $$a_0\ne 0$$, $$a_n\ne 0$$, then the only conceivable rational roots of $$P(x)$$ are of the form $$\dfrac{a}{b}$$, where $$a|a_0$$, $$b|a_n$$. However there are some cases that for $$a|a_0$$, $$b|a_n$$, $$\dfrac{a}{b}$$ migh not be a root of $$P(x)$$.

Are there any general criterion so that $$P(x)$$ doesn't have a rational root?

• en.wikipedia.org/wiki/Rational_root_theorem Commented Oct 25, 2018 at 20:50
• What are you hoping for here? The rational root theorem gives you a good way to test a given polynomial. Note: this has little to do with combinatorics or number theory.
– lulu
Commented Oct 25, 2018 at 20:50
• @lulu It has much to do with number theory. Commented Oct 25, 2018 at 20:55

In general, if we have a polynomial $$P(x)$$ with integer coefficients, where $$P(x)=a_0x^n+\cdots +a_n$$, where $$a_0\ne 0$$, $$a_n\ne 0$$, then the only conceivable rational roots of $$P(x)$$ are of the form $$\dfrac{a}{b}$$, where $$a$$ is a divisor (possibly negative) of $$a_n$$ and $$b$$ is a positive divisor of $$a_0$$.
• Thanks for your answer. However, I want to know when $P(x)$ doesn't have a rational root. If $a|a_n$ and $b|a_0$, $\frac{a}{b}$ might not be a root of $P(x)$ Commented Oct 25, 2018 at 21:00