HMMT 二月 2008 · 代数 · 第 5 题
HMMT February 2008 — Algebra — Problem 5
题目详情
英文原题
- [ 5 ] Let f ( x ) = x + x + 1. Suppose g is a cubic polynomial such that g (0) = − 1, and the roots of g are the squares of the roots of f . Find g (9). n
解析
英文解析
- [ 5 ] Let f ( x ) = x + x + 1. Suppose g is a cubic polynomial such that g (0) = − 1, and the roots of g are the squares of the roots of f . Find g (9).
Answer: 899 Let a, b, c be the zeros of f . Then f ( x ) = ( x − a )( x − b )( x − c ). Then, the roots of
2 2 2 2 2 2
g are a , b , c , so g ( x ) = k ( x − a )( x − b )( x − c ) for some constant k . Since abc = − f (0) = − 1, we
2 2 2
have k = ka b c = − g (0) = 1. Thus,
2 2 2 2 2 2 2
g ( x ) = ( x − a )( x − b )( x − c ) = ( x − a )( x − b )( x − c )( x + a )( x + b )( x + c ) = − f ( x ) f ( − x ) .
Setting x = 3 gives g (9) = − f (3) f ( − 3) = − (31)( − 29) = 899. n