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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
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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

3 Answers 3

up vote 5 down vote accepted

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 –  B. 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
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@TheAlchemist, then edit your post in addition to apologizing. –  J. M. May 7 '13 at 12:35

I believe that all polynomials have rational roots apart from Galios and Abel, hyperoprations will not help, the inversed function must use the the same operation and its inverses of source function.

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Difficult to understand. Needs some proof-reading, punctuation, and elucidation. –  bubba May 7 '13 at 12:20

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