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Message: <blockquote data-quote="imhotep" data-source="post: 30125161" data-attributes="member: 562115">ශේෂ ප්‍රමේය නැවත නැවත පාවිච්චි කරන්නෙ... This is exactly I asked before what's meant by this. I am not aware of such a thing. Sorry.To find the remainder there a only a few ways,  standard long division, then synthetic division and finally the remainder theorem. (there are others like using the Taylor series which can generalize the remainder when divided by (x-a)^m, but they are beyond the scope of A/L)The remainder theorem states....Let P(x) be a polynomial of degree n. If P(x) is divided by x-a , where 'a' is a real number then the remainder is P(a)Note that the divisor has be linear.The remainder theorem can be extended to quadratic divisorsLet P(x) be a polynomial of degree n.  If P(x) is divided by (x-a)(x-b), where both a & b are real, the remainder is of the formR(x) = r1 x + r0 where both r0 & r1 are constants and found by solving P(a) = r1 a + r2 & P(b) = r1 b + r2Note - This doesn't work when the divisors are the same... (x-a)(x-a)  or (x-a)^2Hence AFAIK the derivative has to be used to solve for r0 and r1.PS: If anyone knows about "ශේෂ ප්‍රමේය නැවත නැවත පාවිච්චි කරන්නෙ" please let us know.  [USER=400100]@pdn.ac.lk[/USER]  any ideas?</blockquote>

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