# How can I prove that a polynomial has no rational roots?

Let say I have a polynomial with n as the maximum degree.

Say: $a_1x^n+a_2x^{n-1}+...+a_n$

How can I prove that it does not have any rational roots?

For example: If we have polynomial with degree 2, We can determine it has no roots if $b^2-4ac<0$, But when the degree is higher than 2 what can we do?

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If $n$ is odd it will definitely have a real root. Only if $n$ is even is it possible for there to be no real roots. – Milind May 7 '13 at 10:30
Since OP mentioned the condition $b^2-4ac<0$ as a condition for no roots, I assumed he meant real roots. – Milind May 7 '13 at 10:34
Sorry, I meant rational roots. – StationaryTraveller May 7 '13 at 10:34
Just a little comment on your notation, if you really want to start with $a_1 x^n$, then the last term should be $a_{n+1}$. – Andreas Caranti May 7 '13 at 10:50
math.stackexchange.com/questions/383186/…; may be helpful – Tao May 7 '13 at 10:58

We can use the rational root test. This test states that if an integer-coefficient polynomial $f(x) = a_n x^n + \cdots + a_0$ has a rational root $r = p/q$ and the root is in lowest terms, i.e. $\gcd(p,q)=1,$ then it follows that $$q\mid a_n,\ \ p\mid a_0.$$ Other things we can do include Eisenstein's Criterion, which will show that the polynomial is irreducible, hence doesn't have a root, or we can also use modular arithmetic. If we look at the polynomial in $\Bbb Z/p\Bbb Z$ for some prime, is it irreducible? If the answer is yes, then the poly. is irreducible over the integers, too.

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What do you mean by "For $n$" ? – Arnaud May 7 '13 at 10:47
Nice Clayton. +1 – Babak S. May 8 '13 at 14:10

If $n$ is odd it will definitely have a real root. Only if $n$ is even is it possible for there to be no real roots.

One method when $n$ is even is to determine the points of minima of the polynomial $p(x)$ using calculus. If $a_1>0$ and the minima of $p$ are greater than zero, then $p$ cannot have any roots.

EDIT: For testing of integer roots, take a look at the rational root test.

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Thanks for the answer. But i meant integer roots. Sorry for misleading. – StationaryTraveller May 7 '13 at 10:35
@TheAlchemist, then edit your post in addition to apologizing. – J. M. May 7 '13 at 12:35