HMMT 十一月 2011 · 冲刺赛 · 第 21 题

HMMT November 2011 — Guts Round — Problem 21

[ 10 ] Let P ( x ) = x + 2 x − 13 x − 14 x + 24 be a polynomial with roots r , r , r , r . Let Q be the 1 2 3 4 2 2 2 2 4 quartic polynomial with roots r , r , r , r , such that the coefficient of the x term of Q is 1. Simplify 1 2 3 4 2 the quotient Q ( x ) /P ( x ), leaving your answer in terms of x . (You may assume that x is not equal to any of r , r , r , r ). 1 2 3 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TH 4 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND Round 8

解析

[ 10 ] Let P ( x ) = x + 2 x − 13 x − 14 x + 24 be a polynomial with roots r , r , r , r . Let Q be the 1 2 3 4 2 2 2 2 4 quartic polynomial with roots r , r , r , r , such that the coefficient of the x term of Q is 1. Simplify 1 2 3 4 2 the quotient Q ( x ) /P ( x ), leaving your answer in terms of x . (You may assume that x is not equal to any of r , r , r , r ). 1 2 3 4 4 3 2 Answer: x − 2 x − 13 x + 14 x + 24 We note that we must have 2 2 2 2 2 2 2 2 2 2 2 2 2 Q ( x ) = ( x − r )( x − r )( x − r )( x − r ) ⇒ Q ( x ) = ( x − r )( x − r )( x − r )( x − r ) 1 2 3 4 1 2 3 4 . Since P ( x ) = ( x − r )( x − r )( x − r )( x − r ) , we get that 1 2 3 4 2 Q ( x ) /P ( x ) = ( x + r )( x + r )( x + r )( x + r ) . 1 2 3 4 2 4 Thus, Q ( x ) /P ( x ) = ( − 1) P ( − x ) = P ( − x ), so it follows that 2 4 3 2 Q ( x ) /P ( x ) = x − 2 x − 13 x + 14 x + 24 .