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Let $R$ be a commutative ring with unity, $a\in R$, $f(x) \in R[x]$. Then $a$ is a zero of $f$ iff $x-a$ is a factor of $f$.

Solution: If $a$ is a zero of $f$, then by division algorithm, we can write $f(x) = (x-a)g(x) + r(x)$ such $\deg (x-a) > \deg r$ or $\deg r = 0$ and $g(x) \in R[x]$. But notice $$r(x) = f(x) - g(x)(x-a) \Rightarrow r(a) = 0$$ Therefore $f(x) = g(x)(x-a)$.

the other direction is trivial.

Is this approach correct? can someone give me feedback?

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It's correct. But (unlike with fields) you can't conclude from this that a polynomial of degree n has at most n roots. –  Ted Nov 18 '12 at 6:21
    
Can you explain why $r(a) = 0 \implies r(x) = 0$? –  Vectk Nov 18 '12 at 8:43
    
When you write "using the division algorithm" it makes me suspect. I hope you know that division of polynomials work out in this case because $x-a$ is monic. Since $R$ is a commutative ring, and not a field, there is no division algorithm in general, but only when the polynomial you are dividing by is monic. –  Rankeya Nov 18 '12 at 8:47
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up vote 0 down vote accepted

Your $r(x)$ notation makes things look needlessly complicated : here $r(x)$ is just the constant $p(a)$.

If $$p(x)=\sum_{k=0}^n p_k x^k $$, then

$$ g(x)=\frac{p(x)-p(a)}{x-a}=\sum_{k=0}^n p_k\bigg(\frac{x^k-a^k}{x-a}\bigg) =\sum_{k=0}^n p_k(x^{k-1}+ax^{k-2}+a^2x^{k-3}+\ldots +a^{k-2}x+a^{k-1}) $$

The equality $p(x)=p(a)+(x-a)g(x)$ can also be seen as a Taylor expansion of degree $0$.

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