0
$\begingroup$

This question already has an answer here:

Let $a\in R$. How to show that if a polynomial $p(x)\in R[X]$ where $R$ is an integral domain is not divisible if and only if $p(x+a)$ is not divisible?

$\endgroup$

marked as duplicate by Bill Dubuque divisibility Feb 12 at 17:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Does "not divisible" mean "irreducible"? $\endgroup$ – Bill Dubuque Feb 12 at 17:08
  • $\begingroup$ @BillDubuque As far as I understand, yes! $\endgroup$ – Dole Feb 12 at 17:11
  • $\begingroup$ See also here. $\endgroup$ – Bill Dubuque Feb 12 at 17:17
2
$\begingroup$

Hint: $p(x) = q(x)r(x)$ if, and only if, $p(x + a) = q(x+a)r(x+a)$ (that is, $p(x)$ is reducible if and only if $p(x+a)$ is reducible). Can you fill the gaps?

$\endgroup$

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