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$A(x) , B(x) , C(x) , R_1(x) $ and $R_2(x)$ are polynomials.

Consider $R_1(x) := rem(A(x),C(x)) $ and $R_2(x) := rem(B(x),C(x))$ .

Is it true that $rem(A(x)B(x) \ ,\ C(x)) = rem(R_1(x)R_2(x) \ , \ C(x))$ ? If it's true prove it.

My try : I did it in general form but didn't get any result.

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    $A=CQ_1+R_1$ and $B=CQ_2+R_2 \implies AB=C(Q_1B+R_1Q_2)+R_1R_2$. – 2017-01-23
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    @JonasMeyer Yes but it isn't same as $rem(A(x)B(x) \ ,\ C(x)) = rem(R_1(x)R_2(x) \ , \ C(x))$ – 2017-01-23
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    Related, isn't it? – 2017-01-23
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    @JonasMeyer Yes , it is related but I can't deduct it from your expression. If you can prove it , please write it – 2017-01-23
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    @JonasMeyer Why we can't say "$rem(A(x)B(x) \ ,\ C(x)) = R_1(x)R_2(x)$" – 2017-01-23

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If $P_1$ and $P_2$ are polynomials such that $C$ divides $P_1-P_2$, then $P_1$ and $P_2$ have the same remainder when dividing by $C$.

To see this, we assume $P_1-P_2=CP$ for some polynomial $P$, and with polynomial division obtain $P_1=Cq_1+r_1$, $P_2=Cq_2+r_2$. Then $P_1-P_2=C(q_1-q_2)+r_1-r_2$. From $CP=C(q_1-q_2)+r_1-r_2$ we can conclude that $r_1-r_2$ is a multiple of $C$, which implies that $r_1-r_2=0$ (because $r_1-r_2$ can't have degree as large as $C$).

This applies with $P_1=AB$ and $P_2=R_1R_2$ because if polynomial division yields $A =CQ_1+R_1$ and $B=CQ_2+R_2$, then $AB-R_1R_2=C(Q_1B+R_1Q_2)$.

A question I see added in a comment is, why can't we say that the remainder of $AB$ when dividing by $C$ equals $R_1R_2$? The reason is that $R_1R_2$ might have degree as large or larger than $C$. Example: $A=B=x^2+x$, $C=x^2$, $R_1R_2=x^2$, but the remainder of $AB$ when dividing by $C$ is $0$.

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    Can you explain first part of your answer ? – 2017-01-23
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    @S.H.W. Can you ask a more specific question? – 2017-01-23
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    What do you mean by "If $P_1$ and $P_2$ are polynomials such that $C$ divides $P_1-P_2$ , then $P_1$ and $P_2$ have the same remainder when dividing by $C$" ? – 2017-01-23
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    @S.H.W. I'm having a hard time guessing which part of that statement is unclear, or I'd try to helpfully reword it. The following paragraph, which provides a proof, also provides more detailed context of what is being asserted. I'll just expand some: It's saying that if $P_1-P_2$ is a multiple of $C$, which means $P_1-P_2=CP$ for some polynomial $P$, then $P_1$ and $P_2$ give the same remainder when dividing by $C$, which means if $P_1=Cq_1+r_1$ and $P_2=Cq_2+r_2$ is the result of poly long division, then $r_1=r_2$. Note $P_1=CP+P_2$, just like $AB=C(Q_1B+R_1Q_2)+R_1R_2$. – 2017-01-23
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    The latter equation is the motivation for stating and proving the more general result. – 2017-01-23
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    Is $C$ a number ? – 2017-01-23
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    No, every letter symbol I wrote is a polynomial. Read $A$ as $A(x)$, $B$ as $B(x)$, $C$ as $C(x)$, etc. Since only polynomials were used I had no need to distinguish different types of objects with the "$x$", and I wanted to save typing and have a less cluttered answer. E.g., notice in the example at the end of the answer that $C=x^2$. – 2017-01-23
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    Every thing is awesome and your answer is beautiful but when we say $P_1 - P_2 = CP$ is it in general ? I think it is not true for all $P_1$ and $P_2$ – 2017-01-23
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    @S.H.W: The statement at the beginning is an implication: If the hypothesis holds, then the conclusion holds. It is totally general *conditional* on whether the hypothesis that $P_1-P_2=CP$ holds for some polynomial $P$. If that is the case, then necessarily $P_1$ and $P_2$ give the same remainder. Now it can be applied to each $P_1$ and $P_2$ that do satisfy the hypothesis, and this holds for $P_1=AB$ and $P_2=R_1R_2$. – 2017-01-23
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    Okay , Thanks a lot – 2017-01-23