close
標題:
Maths - Polynonmial
When a certain polynonmial f(x) is divided by (x-1)(3x+2), the remainder is px+q. When f(x) is divided by x-1, the remainder is -4 ; when f(x) is divided by 3x+2, the remainder is 6. Find the values of p and q.
最佳解答:
Suppose f ( x ) = ( x - 1 )( 3x + 2 )Q ( x ) + px + q, f ( 1 ) = -4 ( 1 - 1 )( 3 + 2 )Q ( 1 ) + p + q = -4 p + q = -4 --- ( 1 ) f ( - 2 / 3 ) = 6 ( - 2 / 3 - 1 )( - 2 + 2 )Q ( - 2 / 3 ) - 2p/3 + q = 6 q = 6 + 2p/3 --- ( 2 ) Put ( 2 ) into ( 1 ), p + 6 + 2p/3 = -4 5p/3 = -10 p = -6 q = 6 + 2(-6)/3 = 2 Hence p = -6, q = 2.
其他解答:
Maths - Polynonmial
此文章來自奇摩知識+如有不便請留言告知
發問:When a certain polynonmial f(x) is divided by (x-1)(3x+2), the remainder is px+q. When f(x) is divided by x-1, the remainder is -4 ; when f(x) is divided by 3x+2, the remainder is 6. Find the values of p and q.
最佳解答:
Suppose f ( x ) = ( x - 1 )( 3x + 2 )Q ( x ) + px + q, f ( 1 ) = -4 ( 1 - 1 )( 3 + 2 )Q ( 1 ) + p + q = -4 p + q = -4 --- ( 1 ) f ( - 2 / 3 ) = 6 ( - 2 / 3 - 1 )( - 2 + 2 )Q ( - 2 / 3 ) - 2p/3 + q = 6 q = 6 + 2p/3 --- ( 2 ) Put ( 2 ) into ( 1 ), p + 6 + 2p/3 = -4 5p/3 = -10 p = -6 q = 6 + 2(-6)/3 = 2 Hence p = -6, q = 2.
其他解答:
文章標籤
全站熱搜
留言列表
