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Polynomial Remainder

A polynomial is an expression of the form AnX^n + An-1X^(n-1) + An-2X^(n-2) + ..... + A1X + Ao. Examples are 4x^2 + -3x + 7 and -5x^3 + x^2 - 2x. Let's suppose that a polynomial when divided by x-19 has remainder 99 and the same polynomial when divided by x-99 has remainder 19. What is the remainder when that polynomial is divided by (x-19)(x-99)? Let P(x) be the polynomial. Then we are given that P(x)/(x-19) = Q(x) + 99/(x-19) and P(x)/(x-99) = R(x) + 19/(x-99) where Q(x) and R(x) are polynomials. If we subtract the left and right sides of equation 2 from the left and right sides of equation 1 we see that P(x)/(x-19) - P(x)/(x-99) = Q(x) - R(x) + 99/(x-19) - 19/(x-99). The left side simplifies to [P(x)(x-99) - P(x)(x-19)]/(x-19)(x-99) = (xP(x) - 99P(x) - xP(x) + 19P(x))/(x-19)(x-99) = -80P(x)/(x-19)(x-99). The right side simplifies to Q(x) - R(x) + [99(x-99) - 19(x-19)]/(x-19)(x-99) = Q(x) - R(x) + (99x - 99^2 - 19x + 19^2)/(x-19)(x-99) = Q(x) - R(x) + [80x - (99^2 - 19^2)]/(x-19)(x-99) = Q(x) - R(x) + [80x - (99 - 19)(99 + 19)]/(x-19)(x-99) = Q(x) - R(x) + [80x - 80*118]/(x-19)(x-99) = Q(x) - R(x) + [80(x-118)]/(x-19)(x-99). Our equation has now transformed itself into -80P(x)/(x-19)(x-99) = Q(x) - R(x) + [80(x-118)]/(x-19)(x-99). Dividing both sides by -80 results in P(x)/(x-19)(x-99) = 1/80Q(x) - 1/80R(x) + (-x +118)/(x-19)(x-99). Notice that 1/80Q(x) - 1/80R(x) is a polynomial and therefore we conclude that when P(x) is divided by (x-19)(x-99) the remainder is -x + 118