標題:

maths~~~

發問:

Let P(x)=x3+ax2+bx+15.When P(x) is divided by x-3,the remainder is 66.When P(x) is divided by x+3,the remainder is 0.Find the values of a and b. 更新: Let P(x)=x^3+ax^2+bx+15.When P(x) is divided by x-3,the remainder is 66.When P(x) is divided by x+3,the remainder is 0.Find the values of a and b.

最佳解答:

免費註冊體驗

 

此文章來自奇摩知識+如有不便請留言告知

P(x)=x^3 ax^2 bx 15 By remainder theorem, P(3) = 66 P(-3)= 0 P(3) = 3^3 + a(3)^2 + b(3) +15 27 + 9a + 3b + 15 = 66 9a + 3b = 24 3a + b = 8 ... (1) P(-3) = (-3)^3 + a(-3)^2 + b(-3) +15 -27 + 9a - 3b + 15 = 0 9a - 3b - 12 = 0 3a -b - 4 = 0 3a - b = 4 ... (2) (1) + (2) 6a = 12 a = 2 put b = 2 into (2) 3(2) - b = 4 6 - b = 4 b = 2

其他解答:
arrow
arrow

    jpjzhf9 發表在 痞客邦 留言(0) 人氣()