HMMT 十一月 2011 · 冲刺赛 · 第 21 题
HMMT November 2011 — Guts Round — Problem 21
题目详情
英文原题
- [ 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
the quotient Q ( x ) /P ( x ), leaving your answer in terms of x . (You may assume that x is not equal to 2
any of r , r , r , r ).
1 2 3 4
4 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUNDTH
Round 8
解析
英文解析
- [ 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
the quotient Q ( x ) /P ( x ), leaving your answer in terms of x . (You may assume that x is not equal to 2
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
Q ( x ) /P ( x ) = ( x + r )( x + r )( x + r )( x + r ) .2
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 .