2
$\begingroup$

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

$\endgroup$
  • 8
    $\begingroup$ Yes. That's what the rational root theorem guarantees (given that you can factor the constant term). $\endgroup$ – Qiaochu Yuan Dec 19 '11 at 20:44
  • 2
    $\begingroup$ 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. $\endgroup$ – Bill Dubuque Dec 20 '11 at 0:01
1
$\begingroup$

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 $d^n$

$\endgroup$
  • 2
    $\begingroup$ 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 $\rm\:6\ |\ 3\:.$ $\endgroup$ – Bill Dubuque Dec 19 '11 at 22:12
  • $\begingroup$ @BillDubuque does it work when there are two variables? $\endgroup$ – yiyi Sep 1 '12 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.