MATH 問題

Let P ( x ) = ( x - 2 )( x - 3 ) Q ( x ) + 5.
P ( 2 ) = ( 2 - 2 )( 2 - 3 ) Q ( 2 ) + 5 = 5
P ( 3 ) = ( 3 - 2 )( 3 - 3 ) Q ( 3 ) + 5 = 5
So when P ( x ) is divided by x - 2 or x - 3, the remainder is 5.
Next suppose
P ( x ) = ( x - 1 )( x - 2 )( x - 3 )Q' ( x ) + px2 + qx + r.
P ( 1 ) = p + q + r = 1 --- ( 1 )
P ( 2 ) = 4p + 2q + r = 5 --- ( 2 )
P ( 3 ) = 9p + 3q + r = 5 --- ( 3 )
From ( 1 ) and ( 2 ),
4p + 2q + 1 - p - q = 5
3p + q = 4 --- ( 4 )
From ( 2 ) and ( 3 ),
4p + 2q + r = 9p + 3q + r
q = -5p --- ( 5 )
Put ( 5 ) into ( 4 ),
3p - 5p = 4
p = -2
q = ( - 5 )( - 2 ) = 10
Then,
-2 + 10 + r = 1
r = -7
Hence the required remainder is -2x2 + 10x - 7.

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