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If a polynomial has only integer roots, is it always possible to find a root using the rational roots theorem?

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Yes. That's what the rational root theorem guarantees (given that you can factor the constant term). – Qiaochu Yuan Dec 19 '11 at 20:44
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It's difficult to answer till you clarify precisely what you mean by "find a root". E.g. it could mean anything ranging from a nonconstructive existence proof to a polynomial time root-finding algorithm. – Gone Dec 20 '11 at 0:01

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if $a_nx^n+\cdots+a_0\in\mathbb{Z}[x]$ then every rational root is in the set $\{c/d : c|a_0, d|a_n\}$ as one can see by plugging in $c/d$ and multiplying the whole expression by $d^n$

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Beware that the test requires $\rm\:c/d\:$ in lowest terms, i.e. $\rm\:\gcd(c,d) = 1\:$ else e.g. $\ 6/2\ $ is a root of $\rm\ x - 3\ $ but neither $\rm\ 2\ |\ 1\ $ nor $\rm\:6\ |\ 3\:.$ – Gone Dec 19 '11 at 22:12
@BillDubuque does it work when there are two variables? – yiyi Sep 1 '12 at 15:58

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