用双十字相乘法2x-3y4╳╳3x2y-5得分解式为(2x-3y+4)(3x+2y-5)用待定系数法设分解式为(ax+by+c)(dx+ey+f)=adx^2+(ae+bd)xy+bey^2+(af+cd)x+(bf+ce)y+cf得ad=6,ae+bd=-5,be=-6,af+cd=2,bf+ce=23,cf=-20设a=2,d=3,2e+3b=-5be=-62f+3c=2bf+ce=23cf=20由2e+3b=-5,be=-6,易得b=-3,e=2-3f+2c=23,2f+3c=2解方程组得c=4,f=-5所以分解式为(2x-3y+4)(3x+2y-5)