中四數學,餘式定理

1)Let g ( x ) = ( x - 1 )( 2x + 1 ) Q ( x ) + ax + b where ax + b is the remainder.
g ( x ) = ( 2x2 - x - 1 ) Q ( x ) + ax + b
g ( 1 ) = 5 and so,
a + b = 5 --- ( 1 )
g ( - 1 / 2 ) = 2
- a / 2 + b = 2 --- ( 2 )
( 1 ) - ( 2 ) : 3a / 2 = 3
a = 2
2 + b = 5
b = 3
Hence the required remainder is 2x + 3.
2) Let f ( x ) = ( x - 2 )( x + 1 ) Q ( x ) + px + q.
f ( 2 ) = 7
2p + q = 7 --- ( 1 )
f ( - 1 ) = -2
-p + q = -2 --- ( 2 )
( 1 ) - ( 2 ): 3p = 9
p = 3
-3 + q = -2
q = 1
Hence p = 3, q = 1.
f ( x ) = ( x - 2 )( x + 1 ) Q ( x ) + 3x + 1
f ( x + 3 ) = ( x + 3 - 2 )( x + 3 + 1 ) Q ( x + 3 ) + 3 ( x + 3 ) + 1
= ( x + 1 )( x + 4 ) Q ( x + 3 ) + 3x + 10
Hence the required remainder is 3x + 10.