Polynomials(10 points)

1) Let f ( x ) = ( x + 2 )( x - 5 )Q ( x ) + ax + b where ax + b is the remainder.
f ( - 2 ) = 5
( - 2 + 2 )( - 2 - 5 )Q ( - 2 ) - 2a + b = -5
b = 2a - 5 --- ( 1 )
f ( 5 ) = 9
( 5 + 2 )( 5 - 5 )Q ( 5 ) + 5a + b = 9
b = 9 - 5a --- ( 2 )
Put ( 1 ) into ( 2 ),
2a - 5 = 9 - 5a
7a = 14
a = 2
b = 2 ( 2 ) - 5 = -1
So the required remainder is 2x - 1.
2a) f ( -1 ) = 3 (-1)3 + m(-1)2 - (n)(-1) - 7
= m + n - 10
f ( 3 ) = 3(3)3 + m(3)2 -n(3) - 7
= 9m - 3n + 74
b) Suppose f ( x ) = ( x + 1 )( x - 3 ) + 2x - 4
f ( - 1 ) = m + n - 10
-2 - 4 = m + n - 10
m + n = 4 --- ( 1 )
f ( 3 ) = 9m - 3n + 74
2(3) - 4 = 9m - 3n + 74
9m - 3n = - 72
3m - n = -24 --- ( 2 )
c) ( 1 ) + ( 2 ): m + n + 3m - n = 4 - 24
4m = -20
m = -5
3 ( - 5 ) - n = -24
n = 9
So m = -5, n = 9.



2008-03-28 16:41:12 補充：
In b), it should be: Suppose f ( x ) = ( x + 1 )( x - 3 )Q ( x ) + 2x - 4

1.) when a polynomial f(x) is divided by x-5 , the remainder is 9.
when it is divided by x+2, the remainder is -5.
Find the remainder when f(x) is divided by(x-5)(x+2).