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Let's say I have some function of $x$ and an unknown integer $d$, given by

$f(x) = x^2 + d^2 + dx$

and I want to see if it's divisible by something like $x-d$;as I understand it, we could apply the polynomial remainder theorem and find the remainder, $r = f(d) = 3d^2$. Now, I believe I can then argue that this is zero if $d=0$, which is obviously true. But I also can see that this isn't sufficient alone; for example, if $x-d=1$, then $x-d$ will divide always.

My question is thus, can I say that $x-d$ is a factor of $f(x)$ only when EITHER $d=0$ or $x-d=1$, or are there another situations I'm missing?

DRG
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    It doesn't make sense to say that $x - d = 1$; you're considering this function as $x$ ranges over all of its possible values, and $d$ is unknown but fixed. – Qiaochu Yuan Oct 03 '20 at 19:35
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    What Qiaochu Yuan commented. However, I upvoted your query, for your good presentation and work, and for stretching your intuition by questioning everything. – user2661923 Oct 03 '20 at 19:38

1 Answers1

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When we speak about division over polynomials, it means $p(x)|q(x)$ for any $x$, as @QiaochuYuan said. If you consider $x-d=1$, it means $x$ is no more free variable and is $x=d+1$, so $f(x)=f(d+1)$, and therefore both become constants, and you are speak about division over constant integers! Notice in polynomials, there is no necessity to have integer coefficients for speaking about division.

Ali Ashja'
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