Polynomial and RemainderDate: 02/19/98 at 15:23:36 From: George Freeman Subject: Division of unkown polynomials An unknown polynomial f(x) of degree 37 yields a remainder of 1 when divided by x-1, a remainder of 3 when divided by x-3, a remainder of 21 when divided by x-5. Find the remainder when f(x) is divided by (x-1)(x-3)(x-5). Date: 02/19/98 at 16:53:05 From: Doctor Rob Subject: Re: Division of unkown polynomials Okay! There are a couple of ways of solving this. The first method uses the Remainder Theorem: f(x) = (x-a)*g(x) + f(a). The first condition says that f(x) = (x-1)*g(x) + 1. The second condition says that f(x) = (x-1)*g(x) + 1 = (x-3)*h(x) + 3. Substituting in x = 1 tells you that 1 = -2*h(1) + 3, so h(1) = 1, so h(x) = (x-1)*k(x) + 1. Then f(x) = (x-3)*[(x-1)*k(x) + 1] + 3 = (x-1)*(x-3)*k(x) + x. The third condition says that f(x) = (x-5)*m(x) + 21. Substituting in x = 5 tells you that 21 = (5-1)*(5-3)*k(5) + 5, so k(5) = 2, so k(x) = (x-5)*n(x) + 2. Then f(x) = (x-1)*(x-3)*[(x-5)*n(x) + 2] + x, = (x-1)*(x-3)*(x-5)*n(x) + 2*(x-1)*(x-3) + x, = (x-1)*(x-3)*(x-5)*n(x) + 2*x^2 - 7*x + 6. Thus the remainder sought is 2*x^2 - 7*x + 6. The second method uses the Chinese Remainder Theorem. f(x) = 1 (mod x-1), f(x) = 3 (mod x-3), f(x) = 21 (mod x-5). First we must find any polynomials a(x), b(x) and c(x) such that a(x)*(x-3)*(x-5) + b(x)*(x-1)*(x-5) + c(x)*(x-1)*(x-3) = 1 a(x) = 1/8, b(x) = -1/4, and c(x) = 1/8 will do. Then the answer is given by f(x) = 1*a(x)*(x-3)*(x-5) + 3*b(x)*(x-1)*(x-5) + 21*c(x)*(x-1)*(x-3) (mod (x-1)*(x-3)*(x-5)), = 1*(1/8)*(x-3)*(x-5) + 3*(-1/4)*(x-1)*(x-5) + 21*(1/8)*(x-1)*(x-3) (mod (x-1)*(x-3)*(x-5)), = x^2*(1/8-3/4+21/8) + x*(-1+9/2-21/2) + (15/8-15/4+63/8) (mod (x-1)*(x-3)*(x-5)), = 2*x^2 - 7*x + 6 (mod (x-1)*(x-3)*(x-5)), and the remainder is 2*x^2 - 7*x + 6. It is comforting that the answers agree! Note that the degree is immaterial, as long as it is at least 2. -Doctor Rob, The Math Forum Check out our web site http://mathforum.org/dr.math/ |
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