f.4 maths~

用戶200412

2009-02-13 1:00 am

Please help me to solve the following questions !!Thx~

1.(A)

Show that x的2次方++2x開方巾x+2)= [x+(開方巾 x+2)]的2次方 - 2

(b)

Hence,solve x的2次方+x+2x開方巾x+2=14

2.

When a polynomial P(X) is divided by x-1 and x+2,the remainder are

1 and 3respectively.Find the remainder whenp(x) is divided by

x的2次方- 1

THX~~

THX~~~~~~~~

THX~~~~~~~~~~~~~~~

回答 (1)

myisland8132

2009-02-13 1:18 am

✔ 最佳答案

1.(A)

Show that x^2+x+2x√(x+2)= [x+√ (x+2)]^2 - 2
Since
[x+√(x+2)]^2 - 2
=x^2+2x√(x+2) + x+2-2
=x^2+x+2x√(x+2)
(b)
Hence,solve x^2+x+2x√(x+2)=14
We have x^2+x+2x√(x+2)=14
[x+√(x+2)]^2 - 2 =14
[x+√(x+2)]^2 = 16
[x+√(x+2)] = 4 or [x+√(x+2)] =-4
√(x+2) = 4-x or √(x+2) =-4-x
x+2=16-8x+x^2 or x+2=16+8x+x^2
x^2-9x+14=0 or x^2+7x+14=0 (rejected)
(x-2)(x-7)=0
x=2 or x=7

2.

When a polynomial P(X) is divided by x-1 and x+1,the remainder are

1 and 3 respectively.Find the remainder when p(x) is divided by x^2- 1
Let P(x)=Q(x)(x^2-1)+Ax+B
Sub x=1 P(1)=A+B=1
Sub x=-1 P(-1)=-A+B=3
Solving we have A = -1 and B = 2
So the remainder is -x + 2

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