問3題數學題~

Question 1:
睇x
LHS = -B/2
RHS = -1
Therefore B/2=1 => B=2
睇constant只係可以做到:
2A²+C=-2
(我諗係你比既information唔夠)

Question 2:
Point 1 and 2.

Question 3:
佢話(x-1)同(x+3)係x³+3x²-x-3既factor，咁即係話x³+3x²-x-3可以比佢地整除。
先將(x-1)同(x+3)整大佢，咁就等於：x²+2x-3
再將x³+3x²-x-3除以x²+2x-3，你會得到 (x+1)
Therefore, (x+1)都係佢個factor
Since: (x-1)(x+1)(x+3)=x³+3x²-x-3

第一同第二條問題唔係幾識睇
只係睇到第三條問題....sor- -

(x-1)(x+3)=x^2+2x-3

___x______+1_______________
x^2+2x-3 ) x^3 + 3x^2 - x + 3
x^3 + 2x^2 - 3x
--------------------------------------------
x^2 + 2x - 3
所以最後一個factor係(x+1)
(x+1)(x-1)(x+3) = x³+3x²-x-3

只會第二

1. x²+x+1/4
=0.25(2x+1)(2x+1)
=0.25(2x+1)²

2. 4k²-12km+9m²
=(2k-3m)(2k-3m)
=(2k-3m)²

3. can not be factorized

2006-12-06 17:56:01 補充：
第一if 2A²-Bx/2 C=6x²-x-2 is an idenity ,then the value of A.B.C2A²=6x²A=3xBx/2=xBx=2xB=2C=-2

2006-12-06 18:00:48 補充：
first (x-1)(x 3)=x² 2x-3x³ 3x²-x-3 / x² 2x-3= x 1(x 1)(x-1)(x 3)=x³ 3x²-x-3so the last factor is x 1