一个问题
1.x^2+xy-2y^2+2x+7y-3
2.x^2+3xy+2y^2+4x+5y+3
参考答案:1.x^2+xy-2y^2+2x+7y-3
用待定系数法，
设x^2+xy-2y^2+2x+7y-3=(x+ay+b)(x+cy+d)
把右边乘开，令同类项的系数相等，解出a,b,c,d
a+c=1..............(1)
ac= -2.............(2)
b+d=2..............(3)
bc+ad=7............(4)
bd= -3.............(5)
结果为x^2+xy-2y^2+2x+7y-3=(x+2y-1)(x-y+3)
2.x^2+3xy+2y^2+4x+5y+3
=(x+y)(x+2y)+3(x+y)+(x+2y)+3
=(x+y)(x+2y+3)+(x+2y+3)
=(x+y+1)(x+2y+3)