HMMT 二月 2014 · 代数 · 第 4 题

HMMT February 2014 — Algebra — Problem 4

Let b and c be real numbers, and define the polynomial P ( x ) = x + bx + c . Suppose that P ( P (1)) = P ( P (2)) = 0, and that P (1) 6 = P (2). Find P (0). 4 3 2

解析

Let b and c be real numbers, and define the polynomial P ( x ) = x + bx + c . Suppose that P ( P (1)) = P ( P (2)) = 0, and that P (1) 6 = P (2). Find P (0). 3 1 Answer: − OR − 1 . 5 OR − 1 Since P ( P (1)) = P ( P (2)) = 0, but P (1) 6 = P (2), it follows that 2 2 P (1) = 1 + b + c and P (2) = 4 + 2 b + c are the distinct roots of the polynomial P ( x ). Thus, P ( x ) factors: 2 P ( x ) = x + bx + c = ( x − (1 + b + c ))( x − (4 + 2 b + c )) 2 = x − (5 + 3 b + 2 c ) x + (1 + b + c )(4 + 2 b + c ) . It follows that − (5 + 3 b + 2 c ) = b , and that c = (1 + b + c )(4 + 2 b + c ). From the first equation, we find 2 b + c = − 5 / 2. Plugging in c = − 5 / 2 − 2 b into the second equation yields − 5 / 2 − 2 b = (1 + ( − 5 / 2) − b )(4 + ( − 5 / 2)) . 1 3 Solving this equation yields b = − , so c = − 5 / 2 − 2 b = − . 2 2 4 3 2