多項式的因式分解

8b:
Shouldn't it be factorisation rather than expansion?

2010-07-17 17:08:24 補充：
1(a) 3a(a+b)-(a-b)(a+b)A. 3a(a+b)-(a-b)(a+b)= (a+b)(3a-(a-b))= (a+b)(3a+a-b)= (a+b)(4a-b)1(b) (4m-n)(2m+n)+(7n-m)(2m+n)A. (4m-n)(2m+n)+(7n-m)(2m+n)= (2m+n)(4m-n+7n-m)= (2m+n)(3m+6n)= 3(2m+n)(m+2n)2(a) (2a+b)(x-y)-b(y-x)A. (2a+b)(x-y)-b(y-x)= (2a+b)(x-y)+b(x-y)= (x-y)(2a+b+b)= (x-y)(2a+2b)= 2(x-y)(a+b)2(b) (5x+3y)(2x-y)-(y-x)(y-2x)A. (5x+3y)(2x-y)-(y-x)(y-2x)= (5x+3y)(2x-y)+(y-x)(2x-y)= (2x-y)(5x+3y+y-x)= (2x-y)(4x+4y)= 4(2x-y)(x+y)3 8x(3次方)-8x(2次方)+4x-4A. 8x^3-8x^2+4x-4= 8x^2(x-1)+4(x-1)= 4(x-1)(2x^2+1)4(a) a^2 bc-a+ab^2 c-b-abc^2 +cA. a^2 bc-a+ab^2 c-b-abc^2 +c= a(abc-1)+b(abc-1)-c(abc-1)= (abc-1)(a+b+c)4(b) 2ax-4ay-4bx+8by+6cx-12cyA. 2ax-4ay-4bx+8by+6cx-12cy= 2a(x-y)-4b(x-y)+6c(x-y)= 2(x-y)(a-2b+3c)5 已知2x^2-5x+1=-1 及xy^2=3 , 求8x^3y^2-20x^2y^2+4xy^2的值A. 8x^3y^2-20x^2y^2+4xy^2= 4xy^2(2x^2-5x+1)= 4*3(-1)= -126 己知x^2-3x+1=2及xy=7 , 求6x^2y-2x^3y-2xy的值A. 6x^2y-2x^3y-2xy= 2xy(3x-x^2-1)= 2xy(-x^2+3x-1)= 2xy(-1)(x^2-3x+1)= 2*7*(-1)(2)= -287(a) 因式分解1+m+n+mnA. 1+m+n+mn= (1+m)+n(1+m)= (1+m)(1+n)7(b) 由此 , 因式分解1+m+n+p+mn+mp+np+mnpA. 1+m+n+p+mn+mp+np+mnp= (1+m)(1+n)+p(1+m+n+mn)= (1+m)(1+n)+p(1+m)(1+n)= (1+m)(1+n)(1+p)8(a) 展開(a+1)(a+2)A. (a+1)(a+2)= a^2+2a+a+2= a^2+3a+28(b) 由此 , 展開(x^2+3x)^2+3(x^2+3x)+2A. Let (x^2+3x) be a.So, (x^2+3x)^2+3(x^2+3x)+2= a^2+3a+2= (a+1)(a+2)= (x^2+3x+1)(x^2+3x+2)= (x^2+3x+1)(x+2)(x+1)9(a) 展開(x+2)(2x-1)A. (x+2)(2x-1)= 2x^2-x+4x-2= 2x^2+3x-29(b) 展開(x+2)(2x^2+x-1)A. (x+2)(2x^2+x-1)= 2x^3+x^2-x+4x^2+2x-2= 2x^3+5x^2+x-29(c) 由此 , 因式分解2x^3+7x^2+4x-4A. (2x^2+3x-2)+(2x^3+5x^2+x-2)= 2x^3+x^2+4x-4So, 2x^3+7x^2+4x-4= (2x^2+3x-2)+(2x^3+5x^2+x-2)= (x+2)(2x-1)+(x+2)(2x^2+x-1)= (x+2)(2x-1+2x^2+x-1)= (x+2)(2x^2+3x-2)9(d) 利用(c)的結果 , 找出2736的所有質因數A. Sorry, I don’t know the answer2736 = 2^4*3^2*1910(a) 展開(x-1)(2x+1)A. (x-1)(2x+1)= 2x^2+x-2x-1= 2x^2-x-110(b) 展開x(x-1)(2x+1)A. x(x-1)(2x+1)= x(2x^2-x-1)= 2x^3-x^2-x10(c) 由此 , 因式分解2x^3-3x^2+1A. 2x^3-3x^2+1= (2x^3-x^2-x)-(2x^2-x-1)So, 2x^3-3x^2+1= (2x^3-x^2-x)-(2x^2-x-1)= x(x-1)(2x+1)-(x-1)(2x+1)= (x-1)(2x+1)(x-1)= (2x+1)(x-1)^2

2010-07-17 17:09:22 補充：
1(a) 3a(a+b)-(a-b)(a+b)

A. 3a(a+b)-(a-b)(a+b)
= (a+b)(3a-(a-b))
= (a+b)(3a+a-b)
= (a+b)(4a-b)

1(b) (4m-n)(2m+n)+(7n-m)(2m+n)

A. (4m-n)(2m+n)+(7n-m)(2m+n)
= (2m+n)(4m-n+7n-m)
= (2m+n)(3m+6n)
= 3(2m+n)(m+2n)

2010-07-17 17:09:48 補充：
2(a) (2a+b)(x-y)-b(y-x)

A. (2a+b)(x-y)-b(y-x)
= (2a+b)(x-y)+b(x-y)
= (x-y)(2a+b+b)
= (x-y)(2a+2b)
= 2(x-y)(a+b)

2(b) (5x+3y)(2x-y)-(y-x)(y-2x)

A. (5x+3y)(2x-y)-(y-x)(y-2x)
= (5x+3y)(2x-y)+(y-x)(2x-y)
= (2x-y)(5x+3y+y-x)
= (2x-y)(4x+4y)
= 4(2x-y)(x+y)

2010-07-17 17:10:12 補充：
3 8x(3次方)-8x(2次方)+4x-4

A. 8x^3-8x^2+4x-4
= 8x^2(x-1)+4(x-1)
= 4(x-1)(2x^2+1)

4(a) a^2 bc-a+ab^2 c-b-abc^2 +c

A. a^2 bc-a+ab^2 c-b-abc^2 +c
= a(abc-1)+b(abc-1)-c(abc-1)
= (abc-1)(a+b+c)

4(b) 2ax-4ay-4bx+8by+6cx-12cy

A. 2ax-4ay-4bx+8by+6cx-12cy
= 2a(x-y)-4b(x-y)+6c(x-y)
= 2(x-y)(a-2b+3c)

2010-07-17 17:10:24 補充：
5 已知2x^2-5x+1=-1 及xy^2=3 ,
求8x^3y^2-20x^2y^2+4xy^2的值

A. 8x^3y^2-20x^2y^2+4xy^2
= 4xy^2(2x^2-5x+1)
= 4*3(-1)
= -12

6 己知x^2-3x+1=2及xy=7 , 求6x^2y-2x^3y-2xy的值

A. 6x^2y-2x^3y-2xy
= 2xy(3x-x^2-1)
= 2xy(-x^2+3x-1)
= 2xy(-1)(x^2-3x+1)
= 2*7*(-1)(2)
= -28

2010-07-17 17:10:37 補充：
7(a) 因式分解1+m+n+mn

A. 1+m+n+mn
= (1+m)+n(1+m)
= (1+m)(1+n)

7(b) 由此 , 因式分解1+m+n+p+mn+mp+np+mnp

A. 1+m+n+p+mn+mp+np+mnp
= (1+m)(1+n)+p(1+m+n+mn)
= (1+m)(1+n)+p(1+m)(1+n)
= (1+m)(1+n)(1+p)

2010-07-17 17:10:47 補充：
8(a) 展開(a+1)(a+2)

A. (a+1)(a+2)
= a^2+2a+a+2
= a^2+3a+2

8(b) 由此 , 展開(x^2+3x)^2+3(x^2+3x)+2

A. Let (x^2+3x) be a.
So, (x^2+3x)^2+3(x^2+3x)+2
= a^2+3a+2
= (a+1)(a+2)
= (x^2+3x+1)(x^2+3x+2)
= (x^2+3x+1)(x+2)(x+1)

2010-07-17 17:11:14 補充：
9(a) 展開(x+2)(2x-1)

A. (x+2)(2x-1)
= 2x^2-x+4x-2
= 2x^2+3x-2

9(b) 展開(x+2)(2x^2+x-1)

A. (x+2)(2x^2+x-1)
= 2x^3+x^2-x+4x^2+2x-2
= 2x^3+5x^2+x-2

2010-07-17 17:11:39 補充：
9(c) 由此 , 因式分解2x^3+7x^2+4x-4

A. (2x^2+3x-2)+(2x^3+5x^2+x-2)
= 2x^3+x^2+4x-4

So, 2x^3+7x^2+4x-4
= (2x^2+3x-2)+(2x^3+5x^2+x-2)
= (x+2)(2x-1)+(x+2)(2x^2+x-1)
= (x+2)(2x-1+2x^2+x-1)
= (x+2)(2x^2+3x-2)

9(d) 利用(c)的結果 , 找出2736的所有質因數

A. Sorry, I don’t know the answer
2736 = 2^4*3^2*19

2010-07-17 17:11:45 補充：
10(a) 展開(x-1)(2x+1)

A. (x-1)(2x+1)
= 2x^2+x-2x-1
= 2x^2-x-1

10(b) 展開x(x-1)(2x+1)

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

10(c) 由此 , 因式分解2x^3-3x^2+1

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

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

2010-07-17 17:29:19 補充：
格式問題(其實我也不知道...
以前也沒有出現這情況)

2011-08-01 17:07:18 補充：
To 013:

Thanks for your reminding.

1(a) 3a(a+b)-(a-b)(a+b)

A. 3a(a+b)-(a-b)(a+b)
= (a+b)(3a-(a-b))
= (a+b)(3a-a+b)
= (a+b)(2a+b)

It definitely is my mistake...

我想問下1a點解(a+b)(3a-(a-b))= (a+b)(3a+a-b)
唔係應該(a+b)(3a-a+b) 咩??
我數學好屎,如係本人自己既問題請大家見諒!!

Why's your answer so messy??