HMMT 二月 2003 · 代数 · 第 10 题

英文解析

10. Suppose P ( x ) is a polynomial such that P (1) = 1 and
    P (2 x ) 56 = 8 −
    P ( x + 1) x + 7
    for all real x for which both sides are defined. Find P ( − 1).
    Solution: − 5 / 21
    Cross-multiplying gives ( x + 7) P (2 x ) = 8 x p ( x + 1). If P has degree n and leadingncoefficient c , then the leading coefficients of the two sides are 2 c and 8 c , so n = 3.
    Now x = 0 is a root of the right-hand side, so it’s a root of the left-hand side, sothat P ( x ) = x q ( x ) for some polynomial Q ⇒ 2 x ( x + 7) Q (2 x ) = 8 x ( x + 1) Q ( x + 1) or
    ( x +7) Q (2 x ) = 4( x +1) Q ( x +1). Similarly, we see that x = − 1 is a root of the left-handside, giving Q ( x ) = ( x + 2) R ( x ) for some polynomial R ⇒ 2( x + 1)( x + 7) R (2 x ) =
    4( x + 1)( x + 3) R ( x + 1), or ( x + 7) R (2 x ) = 2( x + 3) R ( x + 1). Now x = − 3 is a rootof the left-hand side, so R ( x ) = ( x + 6) S ( x ) for some polynomial S .
    At this point, P ( x ) = x ( x +2)( x +6) S ( x ), but P has degree 3, so S must be a constant.
    Since P (1) = 1, we get S = 1 / 21, and then P ( − 1) = ( − 1)(1)(5) / 21 = − 5 / 21. 3