解：f(x)g(x)x1 +x -x2-2x-2 x1 +2x3 -x2-4x -2x+1=qi(x)x4 +x3 - x2-2x- 2x4 -2x2= q2(x)x3 +x2-2x-2r(x)=x3-2xxx3x3 -2x= 43(x)-2x0r(x)=x2 -2:: (f(x),g(x)=x2-2f(x) = g(x)+r(x)且由得(g(x)=(x+1)r(x)+r(x)"x2 -2 = -(x+1)f(x)+(x+2)g(x).区区下81.4最大公因式§1.4 最大公因式 4 3 2 x x x x 4 3 2 + − − -2 2 x x x x + − − 2 4 2 - g x( ) f x( ) 4 3 2 x x x x + − − -2 2 1 1 = q x( ) 3 r x 1 ( ) = x x − 2 x+1 4 2 x x − 2 3 2 x x x + − − 2 2 3 x x − 2 2 r x 2 ( ) = x − 2 2 = q x( ) x 3 = q x( ) 3 x x − 2 0 2  = ( ( ), ( )) 2 f x g x x - 2 x x f x x g x − = − + + + 2 ( 1) ( ) ( 2) ( ). 解: 且由 1 1 2 ( ) ( ) ( ) , ( ) ( 1) ( ) ( ) f x g x r x g x x r x r x  = +   = + + 得