PUMaC 2011 · 代数（B 组） · 第 5 题

PUMaC 2011 — Algebra (Division B) — Problem 5

[ 5 ] A polynomial p can be written as 6 5 4 3 2 p ( x ) = x + 3 x − 3 x + ax + bx + cx + d. Given that all roots of p ( x ) are equal to either m or n where m and n are integers, compute p (2).

解析

Let p ( x ) = ( x − m ) ( x − n ) . Note that k cannot be even, as otherwise the coefficient 5 of x would be even. Hence, by symmetry, there are just two cases to check, where k = 1 5 4 (equivalent to k = 5) and k = 3. For k = 1, checking the coefficients of x and x respectively gives m + 5 n = − 3 and 2 − 3 = 5 mn + 10 n = 5 n ( m + 2 n ) = 5 n ( − 3 − 3 n ) , so 5 n ( n + 1) = 1 which certainly has no integral solutions. For k = 3, we obtain respectively 3 m + 3 n = − 3 = ⇒ m + n = − 1 and 2 2 2 − 3 = 3 m + 3 n + 9 mn = 3(( m + n ) + mn ) = 3(1 + mn ) = ⇒ mn = − 2 . 2 Hence, m and n are the roots to the quadratic q ( x ) = ( x − m )( x − n ) = x − x − 2 = ( x − 2)( x +1), 3 3 3 3 so { m, n } = {− 2 , 1 } . Thus, p ( x ) = ( x − 1) ( x + 2) , so the answer is p (2) = 1 · 4 = 64 .