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This is from Pinter, A Book of Abstract Algebra, p.265.

Given $p(x) \in F[x]$ where $F$ is a field, I would like to show that $p(x)$ divided by $x-c$ has remainder $p(c)$.

This is easy if $c$ is a root of $p$, but I don't see how to prove it if $c$ is not a root.

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    If $\dfrac{p(x)}{x-c}=q(x)+\dfrac{r}{x-c}$, then $p(x)=(x-c)q(x)+r$. Now let $x=c$...2011-12-28

5 Answers 5