jmc

Question

A polynomial p(x) leaves a remainder of -1 when divided by x - 1, a remainder of 3 when divided by x - 2, and a remainder of 4 when divided by x + 3. Let r(x) be the remainder when p(x) is divided by (x - 1)(x - 2)(x + 3). Find r(6).

Accepted Answer

By the Remainder Theorem, p(1) = -1, p(2) = 3, and p(-3) = 4. When p(x) is divided by (x - 1)(x - 2)(x + 3), the remainder is of the form ax^2 + bx + c. Thus, p(x) = (x - 1)(x - 2)(x + 3) q(x) + ax^2 + bx + cfor some polynomial q(x). Setting x = 1, x = 2, and x = -3, we get align* a + b + c &= p(1) = -1, 4a + 2b + c &= p(2) = 3, 9a - 3b + c &= p(-3) = 4. align*Subtracting these equations in pairs, we get align* 3a + b &= 4, 5a - 5b &= 1. align*Solving, we find a = 2120 and b = 1720. Then c = -2910, so r(x) = 2120 x^2 + 1720 x - 2910.Thus, r(6) = 2120 6^2 + 1720 6 - 2910 = 40.