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Note that I am talking about rational roots not rational coefficients. I know that Galois theory can tell you but I want to know if knowing whether all the roots of a polynomial are rational can also tell you. Note also that the roots and their properties are usually unknown, but I'm talking about when you only know that they are rational but don't know their values.

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By definition, a polynomial is solvable if and only if its roots can be expressed using rational numbers, addition/subtraction, multiplication/division, and radicals (squareroot symbols, cuberoot symbols, etc.) If the roots are rational, they certainly can be expressed in the above form.

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    Thank you, that is exactly what I wanted to know. So if you know the roots are rational, then Galois theory can help you find them, right? You are talking about THAT "solvable" aren't you?2012-03-07
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    @Kenny: Well, Galois theory doesn't really give you any information in this case, since the Galois group is trivial. But if the roots are rational, you don't need such powerful tools. See Tib's post.2012-03-07
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    What do you mean by "the Galois group would be trivial?" So even if you don't know the roots but do know they are rational, Galois theory doesn't tell you anything about the roots?2012-03-07
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    The Galois group is the automorphism group of the field extension generated by the roots over the starting field. In this case both fields are $\mathbb{Q}$. There are no automorphisms (other than the identity map) of $\mathbb{Q}$ that fix every element of $\mathbb{Q}$2012-03-07
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    I don't see how that is possible. If Galois theory can help you solve solvable polynomials, and solvable polynomials have roots that can be expressed by functions of rational numbers, why can't it help you solve polynomials with rational roots? Does the root have to include at least on radical or something? Or is it because ALL the roots are rational?2012-03-07
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    Well, if you know the Galois group (which often requires knowing the roots--only in degree 4 and lower do we get a general formula for the roots in terms of the coefficients), then there's a way to deduce the roots of an *irreducible* polynomial. But the irreducible factors of a polynomial with rational roots are all of the form $(qX-r)$, where $q$ and $r$ are rational.2012-03-07
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    When CAN Galois theory help you solve polynomials. It can't help you with irrational roots that can't be expressed in radicals because those polynomials are not solvable. So aren't we left with polynomials with rational roots and irrational roots that CAN be expressed in radicals?2012-03-07
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    Yes. Galois theory also explains, in abstract terms, why some equations are solvable by radical and others aren't. And of course there are myriad applications to number theory, algebraic geometry, group theory, etc.2012-03-07
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    So $x^2 - \pi = 0$ isn't solvable? Or did I miss an implied *"in the rationals?"*2012-03-07
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    @BlueRaja-DannyPflughoeft: It is solvable over $\mathbb{Q}(\pi)$, but this post is about polynomials with rational coefficients.2012-03-07
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You can always find all the rational roots of a polynomial using the rational root theorem. See http://en.wikipedia.org/wiki/Rational_root_theorem

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    (1) This does not answer OP's question about whether solvability equivalent to having all roots rational. (2) OP is interested in polynomials whose *all of its roots are rational* rather than finding *all the rational roots* of a polynomial.2012-03-07
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    @J.D. True, but this does answer OP's follow-up comment. Perhaps it would be better as a comment than a solution, but I don't think it deserves a downvote.2012-03-07
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    @BrettFrankel Well, then this is more of a comment. Answers should address the question title & body.2012-03-07
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    BTW I did not downvote! I only posted my remarks.2012-03-07
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    He didn't ask whether it's _equivalent_ to being solvable; he asked whether it _implies_ that it's solvable. Clearly there are some cases where it's solvable and none of the roots are rational.2012-03-07
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    Both questions will suffice.2012-03-07