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A certain polynomial P(x) , $x\in R$ when divided by $x-a, x-b,x-c$ leaves the remainders a,b,c respectively. Find the remainder when P(x) is divided by $(x-a)(x-b)(x-c)$ is (a,b,c are distinct)

My approach : Remainder theorem : $\Rightarrow P(a) =a, P(b) =b; P(c) =c$

Now as per the question let the required remainder by R(x) , then $P(x) =(x-a)(x-b)(x-c) Q(x) +R(x) $

Where R(x) is a remainder of at most degree 2. Now from here I am not getting anything which helps me to solve this problems.. please help thanks..

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You've come a long way. Note that in your factorization we also have $R(a) = a, R(b) = b, R(c) = c$ (since the first term equals $0$), and $R(x)$ is of degree at most $2$. What does that tell you about the polynomial $R(x) - x$?

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Hint $\ {\rm mod}\,\ \color{#c00}{x\!-\! a_i}\!:\,\ f(x)\equiv a_i\!\iff\! f(x)\equiv x\ $ by $\ \color{#c00}{a_i\equiv x}.\ $ So for distinct $\,a_i\in \{a,b,c\}$

$\ \ x\!-\!a_i\mid f\!-\!a_i\!\!\iff\! x\!-\!a_i\mid f\!-\!x \!\iff\! {\rm lcm}\{x\!-\!a_i\}\!\mid f\!-\!x\!\iff\! (x\!-\!a)(x\!-\!b)(x\!-\!c)\mid f\!-\!x$

Remark $\ $ This is an example of the constant case optimization of the Chinese Remainder Theorem (CRT), i.e. an obvious transformation shows that residues have constant value (i.e. independent of the moduli), so CRT reduces to a simple lcm calculation. You can find many more examples of this optimizations in prior posts.

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  • $\begingroup$ @Anant Thanks, fixed. Bad interaction between my macros and SEs. $\endgroup$ – Bill Dubuque Jun 9 '14 at 19:35
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$$R(x)=mx^2+nx+t\\P(a)=R(a)=a\\P(b)=R(b)=b\\P(c)=R(c)=c\\P(a)-P(b)=m(a^2-b^2)+n(a-b)=a-b\\m(a+b)+n=1\\P(a)-P(c)=m(a+c)+n=1\\m(a+b-a-c)=0\\m(b-c)=0$$ Note: i substracted $m(a+b)+n=1$ and $m(a+c)+n=1$ Since b and c are distinct,$m=0$ than from $m(a+b)+n=1$ we get that $n=1$ and from $R(a)=a$ we get that $a+t=a$ so $t=0$

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