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How am I sure that factoring a root from a polynomial in $R[x]$ remains in $R[x]$?

I am sure I studied this during my Ph.D., but the argument escapes me.

SO, suppose we are looking at $f(x) \in R[x],$ with $r \in R$, and $f(r)=0$. How do we know that $g(x)=f(x)/(x-r)$ is in $R[x]$? What is the general result around this area?

Dean C Wills
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Let $f=a_0+a_1X+\dots+a_nX^n$ with $f(r)=0$. Now consider the polynomial $$\begin{align}\tilde f&=f(X+r)\\&=a_0+a_1(X+r)+\dots+a_n(X+r)^n\\&=b_0+b_1X+\dots+b_nX^n.\end{align}$$ The coefficients $b_i$ can be obtained by multiplying out all the parentheses. This is a polynomial with $\tilde f(0)=0$. But $\tilde f(0)=b_0$, so $b_0=0$ and $$\begin{align}\tilde f&=b_1X+\dots+b_nX^n\\ &=X(b_1+\dots+b_nX^{n-1}).\end{align}$$ Now since $f=\tilde f(X-r)$, we now have

$$f=(X-r)[b_1+\dots+b_n(X-r)^{n-1}],$$

where the polynomial in straight brackets is $g$. Note that all the polynomials and factors involved have coefficients in $R$.

Vercassivelaunos
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    What a cute trick! Did you learn this somewhere or come up with it yourself? – Arkady Jul 02 '22 at 07:14
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    @Arkady I came up with it when I taught my high schoolers about multiplicity of roots, which required factoring the polynomials, but when I didn't have polynomial division at my disposal. It's easier to come up with in the reals, because there's a nice graphical way to interpret that shift by $r$ and $-r$, but it works over any ring. – Vercassivelaunos Jul 02 '22 at 07:22
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 02 '22 at 07:35
  • Thank you for finding the related ; i did not find in search, as they’re not identical. – Dean C Wills Jul 02 '22 at 08:32