用十字相乘法法,把y作为常数,x 做降幂排列.
原式=2x2+(y-4)x+(-y2+5y-6)
=2x2+(y-4)x+[-(y2-5y+6)]
=2x2+(y-4)x+[-(y-2)(y-3)]
作十字分解,如下：
1 y-3
2 -y+2
则：
原式=[1x+(y-3)][2x+(-y+2)]
=(x+y-3)(2x-y+2)
验算,结果=2x2-xy+2x+2xy-y2+2y-6x+3y-6
=2x2+xy-y2+5y-6=题目的式子 无误
2x^2+xy-y^2-4x+5y-6如何因式分解?