Question

用待定系数法解题：2x^2+3xy+y^2-4x-3y+2

要过程，急！！！！！！！！！！！！！！！！！！！

Answer 1

关键“齐二次的分解”
2x^2+3xy+y^2＝（2x+y)(x+y)
设[(2x+y)+n][(x+y)+M]
=（2x+y)(x+y)+n(x+y)+m(2x+y)+mn
=（2x+y)(x+y)+(n+2m)+(n+m)y+mn
对比系数得
n+2m=-4
且m+n=-3
且mn=2
n＝－2，m＝－1
原式＝（2x+y-2)(x+y-1)

Answer 2

原=(2x^2+3xy+y^2)-(4x+3y)+2
=(x+y)(2x+y)-(4x+2y+y)+2
=(x+y)(2x+y)-2(2x+y)-y+2
=(2x+y)(x+y-2)-(y-2)
=(2x+y)(x+y-2)-(x+y-2)+x
=(x+y-2)(2x+y-1)+x