若x^2-3x+2xy+y^2-3y-40=(x+y+m)(x+y+n),求m,n的值分别是多少?利用十字相乘法,

Question

若x^2-3x+2xy+y^2-3y-40=(x+y+m)(x+y+n),求m,n的值分别是多少?
利用十字相乘法,

Answer 1

x^2-3x+2xy+y^2-3y-40=(x+y+m)(x+y+n),x^2+2xy+y^2-3x-3y-40=(x+y+m)(x+y+n),(x+y)^2-3(x+y)-40=(x+y+m)(x+y+n),(x+y+5)(x+y-8)=(x+y+m)(x+y+n),m=5n= -8或者m= -8,n=5