Mathz*因式定理

2007-10-16 5:48 am
1* 設x^2+2x-15可被x+k整除。求k的值。

2* 已知f(x)=ax^3+bx^2-3,其中a、b為常數。若f(x)可被x-1整除,而f(x)除以x+1的餘數為-2。求a、b的值。

3* 若f(x)=ax^3+4x^2+bx-6可被x^2+2x-3整除。求a、b的值。

回答 (2)

2007-10-16 6:21 am
✔ 最佳答案
1) f ( x ) = x^2 + 2x - 15

若f ( x )可被x+k整除

f ( - k ) = 0

( - k )^2 + 2 ( - k ) - 15 = 0

k^2 - 2k - 15 = 0

( k - 5 )( k + 3 ) = 0

k = 5 或 -3

2) f(x)=ax^3+bx^2-3

f ( 1 ) = 0

a ( 1 )^3 + b ( 1 )^2 - 3 = 0

a + b = 3 --- ( 1 )

f ( - 1 ) = -2

a ( - 2 )^3 + b ( - 2 )^2 - 3 = -2

-8a + 4b - 1 = 0

a = ( 4b - 1 ) / 8 --- ( 2 )

代( 2 ) 入(1 ),

( 4b - 1 ) / 8 + b = 3

4b - 1 + 8b = 24

12b = 25

b = 25 / 12

a + 25 / 12 = 3

a = 11 / 12

3) x^2+2x-3 = ( x + 3 )( x - 1 ), 所以

f ( - 3 ) = 0

a ( - 3 )^3 + 4 ( - 3 )^2 + b ( - 3 ) - 6 = 0

-27a + 36 - 3b - 6 = 0

9a + b = 10 --- ( 1 )

f ( 1 ) = 0

a ( 1 )^3 + 4 ( 1 )^2 + b ( 1 ) - 6 = 0

a + b = 2 --- ( 2 )

( 1 ) - ( 2 ) : 8a = 8

a = 1

1 + b = 2

b = 1


參考: My Maths Knowledge
2007-10-20 7:10 am
1) k = 5, -3
2) a = 11 / 12, b = 25 / 12
3) a = b = 1


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