∵x^2+2xy-3y^2 = (x+3y)(x-y)
∴设原式=(x+3y+A)(x-y+B)=x^2+2xy-3y^2+(A+B)x+(3B-A)y+AB
∴A+B=3
AB=2
3B-A=1
A=2
B=1
原式=(x+3y+2)(x-y+1)