Question

几道因式分解的题

y(y+1)(x^2+1)+x(2y^2+2y+1)

6x^2-5xy-6y^2-2xz-23yz-20z^2

2x^2-7xy+6y^2+2x-y-12

Answer 1

1.
y(y+1)(x^2+1)+x(2y^2+2y+1)
＝y(y+1)x^2+y(y+1)+x(2y^2+2y+1)
＝y(y+1)x^2+(2y^2+2y+1)x+y(y+1)——十字相乘法
＝[yx+y+1][(y+1)x+y]
＝[yx+y+1][yx+x+y]
2.
6x^2-5xy-6y^2-2xz-23yz-20z^2
=(2x-3y)(3x+2y)-2xz-23yz-20z^2 (把前三项用十字相乘法因式分解）

所以设：原式=(2x-3y+az)(3x+2y+bz)

展开得：=(2x-3y)(3x+2y)+(2x-3y)bz+az(3x+2y)+abz^2
=(2x-3y)(3x+2y)+(2b+3a)xz-(3b-2a)yz+abz^2

左右对应相等，则有：

2b+3a=-2
3b-2a=23
ab=--20
解得：a=5,b=-4

所以原式=(2x-3y+5z)(3x+2y-4z)

3.
2x^2-7xy+6y^2+2x-y-12
=(x-2y)(2x-3y)+2x-y-12
=(x-2y)(2x-3y)+6x-4x-9y+8y-12
=(x-2y)(2x-3y)+6x-9y-4x+8y-12
=(x-2y)(2x-3y)+3(2x-3y)-4(x-2y+3)
=(x-2y+3)(2x-3y)-4(x-2y+3)
=(x-2y+3)(2x-3y-4)