Let $R$ be a commutative ring with identity. Consider the polynomial ring $R[x]$. Suppose $f \in R[x]$ and $a \in R$ are such that $f(a) = 0$. Is it true that $f(x) = (x - a)g(x)$ for some $g \in R[x]$?
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1Yes, it is true. What is the constant coefficient of $f\left(x+a\right) $ ? – darij grinberg Dec 09 '15 at 21:46
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That was clever! Thank you very much! – Brisão Dec 09 '15 at 21:56
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If $f(x)=\sum_{k=0}^n a_k x^k$, we can write $$f(x)=f(x)-f(a)=\sum_{k=0}^n a_k(x^k-a^k).$$ Now for each $k\in \mathbf N$, $x^k-a^k$ is divisible by $x-a$: in any commutative ring, $$x^k-a^k=(x-a)(x^{k-1}+x^{k-2}a+\dots+xa^{k-2}+a^{k-1}).$$
Bernard
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