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...