Mathz*餘式定理

1)設f ( x ) = ( x - 2 )( x + 3 ) + 2,

f ( k ) = k^2

( k - 2 )( k + 3 ) + 2 = k^2

k^2 + 3k - 2k - 6 + 2 = k^2

k - 4 = 0

k = 4

2) f (x) = 4x^2 - 3x - 5

f ( x - 1 ) = 4 ( x - 1 )^2 - 3 ( x - 1 ) - 5

= 4 ( x^2 - 2x + 1 ) - 3x + 3 - 5

= 4x^2 - 8x + 4 - 3x - 2

= 4x^2 - 11x + 2

設f ( x - 1 ) = g ( x ),

g ( x ) = 4x^2 - 11x + 2

g ( - 1 ) = 4 ( - 1 )^2 - 11 ( - 1 ) + 2

= 17

所以f(x-1)除以x+1時的餘數是17。

3)設f ( x ) = x^3 + kx^2 - x + 1,

f ( 2 ) = 2^3 + k ( 2 )^2 - 2 + 1

= 4k + 7

f ( 1 ) = 1^3 + k ( 1 )^2 - 1 + 1

= k + 1

已知: f ( 2 ) = 3 f ( 1 )

4k + 7 = 3 ( k + 1 )

4k + 7 = 3k + 3

k = -4

4) 設f ( x ) = x^2 + ax - b, g ( x ) = bx^2 - ax - 1,

f ( - 2 ) = - 4

( - 2 )^2 + a ( - 2 ) - b = -4

2a + b = 8 --- ( 1 )

g ( 2 ) = 1

b( 2 )^2 - a ( 2 ) - 1 = 1

a = 2b - 1 --- ( 2 )

代( 2 )入( 1 ),

2 ( 2b - 1 ) + b = 8

4b - 2 + b = 8

5b = 10

b = 2

a = 2 ( 2 ) - 1 = 3

所以a = 2, b = 3。