Ｈｅｌｐ！Ｆｒａｃｔｏｒｉｚａｔｉｏｎ！Ｆ．３

1)x^3(x-2y)-y^3(y-2x)

=x^4-2x^3y-y^4+2xy^3--------------------將它展開

=x^4-y^4-2x^3y+2xy^3--------------------將它排列好

=[x^4-y^4]+[-2x^3y+2xy^3]--------------排列好

=[(x^2+y^2)(x^2-y^2)]+[-2xy(x^2-y^2)]---------------------因式分解

=(x^2-y^2)(x^2-2xy+y^2)------------------------------將共同的因子抽出來

=[(x^2-y^2)][(x^2-2xy+y^2)]-----------------------排好

=[(x+y)(x-y)][(x-y)^2]----------------------- 左面:因式分解;右面:十字相乘法

=(x-y)^3(x+y)------------------------------因式分解+答案

p^2(2q-1)-4q^2(p-1)-2q+p

=2p^2q-p^2-4pq^2+4q^2-2q+p--------------------展開

=2pq(p-2q)-(p^2-4q^2)+(p-2q)--------------------因式分解

=2pq(p-2q)-(p+2q)(p-2q)+(p-2q)-------------------中間:十字相乘

=(p-2q)(2pq-p-2q+1)-------------------------------因式分解

=(p-2q)(2pq-p-2q+1)------------------------------排好

=(p-2q)[2q(p-1)-(p-1)]-------------------------因式分解

=(p-2q)(p-1)(2q-1)-------------------------因式分解+答案

a(a+b+c)(a-b+c)+b(b+c-a)(c-a-b)

=a(a^2-b^2+2ac+c^2)+b(a^2-b^2-2ac+c^2)-------------展開(煩)

=a^3-ab^2+2a^2c+ac^2+a^2b-2abc-b^3+bc^2----------再展開(好煩)

=(a-b)(a^2+ab+b^2)+(a-b)ab-(a-b)2ac-(a-b)c^2--------因式分解(好多,小心D)

=(a-b)(a^2+2ab+b^2+2ac-c^2)--------------------取(a-b)+答案(爽呀)


4(ab-xy)^2 -(a^2+b^2-x^2-y^2)^2

=(2ab-2xy)^2-(a^2+b^2-x^2-y^2)^2-----------------------展開(=.=)

=(2ab-2xy+a^2+b^2-x^2-y^2)(2ab-2xy-a^2-b^2+x^2+y^2)---再展開(腦袋的轟炸..嗚嗚...)

=[a^2+2ab+b^2-(x^2+2xy+y^2)][x^2-2xy+y^2-(a^2-2ab+b^2)]--將它們排成一個個可十字相乘的形式

=[(a+b)^2-(x+y)^2][(x-y)^2-(a-b)^2]---------------------因式分解

=(a+b+x+y)(a+b-x-y)(x-y+a-b)(x-y-a+b)-----------因式分解+答案(finish)

因式分解未必一定一開始就因式分解先ga!!!

x^3(x-2y)-y^3(y-2x)
=x^4-2x^3y-y^4+2xy^3
=x^4-y^4-2x^3y+2xy^3
=(x^2+y^2)(x^2-y^2)-2xy(x^2-y^2)
=(x^2-y^2)(x^2-2xy+y^2)
=(x+y)(x-y)(x-y)^2
=(x+y)(x-y)^3
p^2(2q-1)-4q^2(p-1)-2q+p
=2p^2q-p^2-4pq^2+4q^2-2q+p
=2pq(p-2q)-(p^2-4q^2)+(p-2q)
=2pq(p-2q)-(p+2q)(p-2q)+(p-2q)
=(p-2q)(2pq-p-2q+1)
=(p-2q)[2q(p-1)-(p-1)]
=(p-2q)(p-1)(2q-1)

2007-10-30 13:23:36 補充：
a(a+b+c)(a-b+c)+b(b+c-a)(c-a-b)=a(a^2-b^2+2ac+c^2)+b(a^2-b^2-2ac+c^2)=a^3-ab^2+2a^2c+ac^2+a^2b-2abc-b^3+bc^2=(a-b)(a^2+ab+b^2)+(a-b)ab-(a-b)2ac-(a-b)c^2=(a-b)(a^2+2ab+b^2+2ac-c^2)

2007-10-30 20:10:21 補充：
4(ab-xy)^2 -(a^2+b^2-x^2-y^2)^2=(2ab-2xy)^2-(a^2+b^2-x^2-y^2)^2=(2ab-2xy+a^2+b^2-x^2-y^2)(2ab-2xy-a^2-b^2+x^2+y^2)=[a^2+2ab+b^2-(x^2+2xy+y^2)][x^2-2xy+y^2-(a^2-2ab+b^2)]=[(a+b)^2-(x+y)^2][(x-y)^2-(a-b)^2]=(a+b+x+y)(a+b-x-y)(x-y+a-b)(x-y-a+b)