HMMT 二月 2003 · 代数 · 第 10 题

HMMT February 2003 — Algebra — Problem 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). 1

解析

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 xP ( x + 1). If P has degree n and leading n coefficient 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, so that P ( x ) = xQ ( 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-hand side, 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 root of 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