(I) 证：若f(x)±0，则x(g(x)+h(x)= f(x)±0,从而 g(x)+h(x)≠0. 于是a(xg (x) + xh(x)) = a(x(g (x)+ h(x)为奇数.但 a(f2(x)为偶数. : x(g(x)+h(x)± f2(x),故 (x)=0,这与已知矛盾.从而 g(x)+h(x)= 0.F81.2一元多项式§1.2 一元多项式 (1) 证：若 f x( ) 0,  则 2 2 2 x g x h x f x ( ( ) ( )) ( ) 0, + =  于是 2 2 2 2  + =  + ( ( ) ( )) ( ( ( ) ( ))) xg x xh x x g x h x 为奇数. 故 f x( ) 0, = 从而 2 2 g x h x ( ) ( ) 0. + = 从而 2 2 g x h x ( ) ( ) 0. +  2 但 ( ( )) f x 为偶数. 这与已知矛盾. 2 2 2  +  x g x h x f x ( ( ) ( )) ( )