f(x)=x^2+2x+3)^3除以x^2+2x+2的餘式為?
答案1
設f(x)=x^8+2x^7-5x^4+4x^3-x^2+1,則f(x)除以x^3-1的餘式為?
答案-3x+5
需詳細過程,謝謝!!
急~~高中數學求餘式?
2014-04-16 11:55 pm
回答 (9)
2014-04-17 12:23 am
✔ 最佳答案
令：f(x)=Q(x) (2x+3)^2 (x-2)^2+a(x-2)(2x+3)^2+b(2x+3)^2+x+6f(2)=49b+8 =57,b=1f(x)=Q(x) (2x+3)^2 (x-2)^2+a(x-2)(2x+3)^2+(2x+3)^2+x+6因為f(x) 除以 (x-2)^2 的餘數= a(x-2)(2x+3)^2+(2x+3)^2+x+6除以 (x-2)^2 的餘數= 78x-99令x^2-4x+4=0 ，則 x^2=4x-4 代入(2x+3)^2=4x^2+12x+9=16x-16+12x+9=28x-7=7(4x-1)a(x-2)(2x+3)^2+(2x+3)^2+x+6= 7a (x-2)(4x-1) +7(4x-1)+x+6= 7a (4x^2-9x+2) +7(4x-1)+x+6= 7a (16x-16-9x+2) +7(4x-1)+x+6= 7a (7x-14) +7(4x-1)+x+67a(7x-14) +7(4x-1)+x+6=78x-997a(7x-14) +7(4x-1)=77x-105a(7x-14) +(4x-1)=11x-15a(7x-14)=7x-14a=1f(x)除以(2x+3)^2 (x-2)^2的餘數：(x-2)(2x+3)^2+(2x+3)^2+x+6
參考: kon
2014-05-30 4:52 pm
2014-05-08 3:25 pm
2014-05-01 4:33 pm
2014-04-29 3:26 pm
2014-04-28 3:14 pm
2014-04-25 2:15 pm
2014-04-24 1:18 pm
2014-04-17 1:39 am
f(x)=(x^2+2x+3)^3除以x^2+2x+2的餘式為?
Sol
f(x)=(x^2+2x+3)^3
=[(x^2+2x+2)+1]^3
=(x^2+2x+2)^3+3(x^2+2x+2)^2+3(x^2+2x+2)+1
So
(x^2+2x+3)^3除以x^2+2x+2的餘式為1
設f(x)=x^8+2x^7－5x^4+4x^3－x^2+1，則f(x)除以x^3－1的餘式為?
Sol
x^3－1=0
x^3=1
So
x^8+2x^7－5x^4+4x^3－x^2+1
=(x^3)^2*x^2+2*(x^3)^2*x－5*(x^3)*x+4*(x^3)－x^2+1
=>1^2*x^2+2*1^2*x－5*(1)*x+4*(1)－x^2+1
=x^2+2x－5x+4－x^2+1
=－3x+5
Sol
f(x)=(x^2+2x+3)^3
=[(x^2+2x+2)+1]^3
=(x^2+2x+2)^3+3(x^2+2x+2)^2+3(x^2+2x+2)+1
So
(x^2+2x+3)^3除以x^2+2x+2的餘式為1
設f(x)=x^8+2x^7－5x^4+4x^3－x^2+1，則f(x)除以x^3－1的餘式為?
Sol
x^3－1=0
x^3=1
So
x^8+2x^7－5x^4+4x^3－x^2+1
=(x^3)^2*x^2+2*(x^3)^2*x－5*(x^3)*x+4*(x^3)－x^2+1
=>1^2*x^2+2*1^2*x－5*(1)*x+4*(1)－x^2+1
=x^2+2x－5x+4－x^2+1
=－3x+5
收錄日期: 2021-04-30 18:35:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140416000010KK03966