jmc

Question

Let P(x) be a polynomial such that when P(x) is divided by x - 19, the remainder is 99, and when P(x) is divided by x - 99, the remainder is 19. What is the remainder when P(x) is divided by (x - 19)(x - 99)?

Accepted Answer

By the Remainder Theorem, P(19) = 99 and P(99) = 19. When P(x) is divided by (x - 19)(x - 99), the remainder must be of the form ax + b. Thus, P(x) = (x - 19)(x - 99) Q(x) + ax + b,for some polynomial Q(x). Setting x = 19 and x = 99, we get align* 19a + b &= P(19) = 99, 99a + b &= P(99) = 19. align*Subtracting the equations, we get 80a = -80, so a = -1. Then -19 + b = 99, so b = 118. Hence, the remainder is -x + 118.