Form 2 Maths

2007-09-28 2:14 am
Factorize the following expressions

1) -20p(3p-2)^2 - 20(3p-2)p - 5p

2)1- x^2 +4xy -4y^2

3) a ^3 - b^3 + 3ab^2 - 3a^2 b

回答 (4)

2007-09-28 2:22 am
✔ 最佳答案
1)
-20p(3p-2)^2 - 20(3p-2)p - 5p
= -20p(9p^2 - 12p + 4) - 20p(3p-2) - 5p
= -180p^3 + 240p^2 - 80p - 60p^2 + 40p - 5p
= -180p^3 + 180p^2 - 45p
= -45p(4p^2 - 4p + 1)
= -45p(2p-1)^2

2)
1 - x^2 + 4xy - 4y^2
= 1 - (x^2 - 4xy + 4y^2)
= 1 - (x-2y)^2
= 1^2 - (x-2y)^2
= (1+x-2y) [ 1 - (x-2y) ] (Be noted that a^2 - b^2 = (a+b)(a-b) )
= (1+x-2y)(1-x+2y)

3)
a^3 - b^3 + 3ab^2 - 3a^2b
= (a-b)(a^2 + ab + b^2) + 3ab(b-a)
= (a-b)(a^2 + ab + b^2) - 3ab(a-b)
= (a-b) [ (a^2 + ab + b^2 - 3ab) ]
= (a-b)(a^2 - 2ab + b^2)
= (a-b)(a-b)^2
= (a-b)^3
參考: I am a Maths tutor
2007-09-29 5:36 am
1)
-20p(3p-2)^2 - 20(3p-2)p - 5p
= -20p(9p^2 - 12p + 4) - 20p(3p-2) - 5p
= -180p^3 + 240p^2 - 80p - 60p^2 + 40p - 5p
= -180p^3 + 180p^2 - 45p
= -45p(4p^2 - 4p + 1)
= -45p(2p-1)^2

2)
1 - x^2 + 4xy - 4y^2
= 1 - (x^2 - 4xy + 4y^2)
= 1 - (x-2y)^2
= 1^2 - (x-2y)^2
= (1+x-2y) [ 1 - (x-2y) ] (Be noted that a^2 - b^2 = (a+b)(a-b) )
= (1+x-2y)(1-x+2y)

3)
a^3 - b^3 + 3ab^2 - 3a^2b
= (a-b)(a^2 + ab + b^2) + 3ab(b-a)
= (a-b)(a^2 + ab + b^2) - 3ab(a-b)
= (a-b) [ (a^2 + ab + b^2 - 3ab) ]
= (a-b)(a^2 - 2ab + b^2)
= (a-b)(a-b)^2
= (a-b)^3
參考: My maths teacher
2007-09-28 2:27 am
1/ -20p(3p-2)^2 - 20(3p-2)p - 5p
= - 5p [ (3p-2)^2 + (3p-2) + 1]
= - 5p ( 3p -2 +1 )^2
= - 5p ( 3p - 1 )^2

2/ 1- x^2 +4xy -4y^2
= 1- [ x^2 - 4xy + (2y)^2 ]
= 1- ( x - 2y )^2
= (1 + x - 2y )(1 - x + 2y)

3/
( a - b )^3
=a^3 - 3a^2b + 3ab^2 - b^3

a ^3 - b^3 + 3ab^2 - 3a^2 b
=a ^3 - 3a^2 b+ 3ab^2 - b^3
=( a - b )^3
參考: myself
2007-09-28 2:23 am
1. -20p(3p - 2)2 – 20(3p - 2)p – 5p

= -5p[4(3p - 2)2 + 4(3p - 2) + 1]

Let x be 3p – 2

i.e. -5p(4x2 + 4x + 1)

= -5p[(2x)2 + 2(2x)(1) + (1)2]

= -5p(2x + 1)2

= -5p[2(3p - 4) + 1]2

= -5p(6p - 7)2


2. 1 – x2 + 4xy – 4y2

= 1 – (x2 – 4xy + 4y2)

= 1 – [x2 – 2x(2y) + (2y)2]

= 1 – (x – 2y)2

= (1)2 – (x – 2y)2

= [1 + (x – 2y)][1 – (x – 2y)]

= (1 + x – 2y)(1 – x + 2y)


3. a3 – b3 + 3ab2 – 3a2b

= (a3 – b3) + (3ab2 – 3a2b)

= (a - b)(a2 + ab + b2) + 3ab(b - a)

= (a - b)(a2 + ab + b2) – 3ab(a - b)

= (a - b)(a2 + ab + b2 – 3ab)

= (a - b)(a2 – 2ab + b2)

= (a - b)(a - b)2

= (a - b)3
參考: Myself~~~


收錄日期: 2021-04-13 19:04:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070927000051KK02456

檢視 Wayback Machine 備份