要設成x2y2+4xy-y2-x2y-x-y+2=(xy+ax+by+c)(xy+dx+ey+f)

x2y2+4xy-y2-x2y-x-y+2=(x^2+ax+by+c)(y^2+dy+ex+f),計算結果矛盾

如果設x2y2+4xy-y2-x2y-x-y+2=(x^2y+axy+bx+cY+d)(y+ex+f)
因為ax.x^2y=ax^3y超過原式的次項,所以a=0
即x2y2+4xy-y2-x2y-x-y+2=(x^2y+ax+by+c)(y+f),計算結果矛盾

總之蠻麻煩的