分解因式 (1)x^2+y^2+6x-2y+8 (2) (x^2+3x+3）（x^2+3x+7
问题描述：
分解因式 (1)x^2+y^2+6x-2y+8 (2) (x^2+3x+3）（x^2+3x+7
分解因式 (1)x^2+y^2+6x-2y+8 (2) (x^2+3x+3）（x^2+3x+7）+4 速回,要大致过程.
答
x^2y^2-x^2-y^2+1=(x^2y^2-2xy+1)-(x^2-2xy+y^2)=(xy-1)^2-(x-y)^2=(xy-x+y-1)(xy+x-y-1)6x^2+2x-5xy-6y^2+23y-20=6x^2-5xy-6y^2+2x+23y-20=(3x+2y)(2x-3y)+2x+23y-20=(3x+2y-5)(2x-3y+4)(x^2+3x+3)(x^2+3x+7)+4
=(x^2+3x+3)[(x^2+3x+3)+4]+4
=(x^2+3x+3)^2+4(x^2+3x+3)+4
=[(x^2+3x+3+2]^2
=(x^2+3x+5)^2x^2-y^2-6x-2y+8
=x^2-6x+9-y^2-2y-1
=(x^2-6x+9)-(y^2+2y+1)
=(x-3)^2-(y-1)^2
=[(x-3)+(y-1)]×[(x-3)-(y-1)]
=(x+y-4)(x-y-2)