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Message: <blockquote data-quote="imhotep" data-source="post: 30114400" data-attributes="member: 562115">Maybe there's some trick. I am not aware of a repeated use of the reminder theorem. My current knowledge of maths is limited mostly to what I taught my son when he sat for A/L , but it's a quite while now.If P(x) is a polynomial with a degree of m, then the remainder when divided by (x-a)^n  (when m >= n)  will be another polynomial of degree n-1. Let's say it's R(x)Then P(x) = (x-a)^n Q(x) + R(x)When n = 1, R(x) will have a degree of 0 (a constant) , let's say r0Then P(x) = (x-a) Q(x) + r0,  with x = a this gives us that P(a) = r0When n = 2, R(x) will have a degree of 1 which is of the form R(x) = r1x + r0Then P(x) = (x-a)^2 Q(x) +(r1x +r0), with x=a this gives us P(a) = r1a + r0Differentiation gives us P'(x) = 2(x-a)Q(x) + (x-a)^2 Q'(x) + r1 This gives us P'(a) = r1. So these set of equations will give you r1 and r0Note:  This is also quite evident from the Taylor series expansion of P(x) if you know about it.P(x) = P(a) + P'(a) (x-a) + P"(a) (x-a)^2 / 2 +........ etc ( can't type more terms in simple text)So the given expression 2x^3+7x^2+13x-3 when divied by x+1 will have a remainder of -11when divided by (x+1)^2 will have a remainder of 5x-6  (calculate r1 and r0 as  mentioned above)when divided ny (x+1)^3 will have a remainder of x^2+7x-5</blockquote>

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