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This seems obvious, but I just can't crack it.

Let K be a field and $F(X) \in K[X]$ be a polynomial. Does $F(a)=0$ for some $a\in K$ imply that F(X) is reducible. Clearly, by the fundamental theorem of algebra, $F(X) = (X-a)G(X)$ for some $G(X) \in \bar{K}[X]$, but does it follow that actually $G(X) \in K[X]$?

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    The answer to the question: NO. A polynomial of degree 1 may have a root but still be irreducible. So: state it more carefully.2012-04-05
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    The title of the question does not really reflect what is being asked.2012-04-05
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    @lhf Yes, it does. In this context "rational" is a common abbreviation for $K$-rational. It is also the better formulation to use in titles, where there is no other reference to $K$.2012-04-05
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    It's obvious that $G(X) \in K(X)$ if you think to think about that. What can you do with that information?2012-04-05

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