Question

因式分解x2-3xy-4y2+4x-y+3

2在后面是平方

Answer 1

算了半天，终于算出来了。哈哈，
x2-3xy-4y2+4x-y+3
=(x+y)(x-4y)+4x-y+3 /将x2-3xy-4y2合并/
=(x+y)(x-4y)+3(x+y)+(x-4y)+3 /将4x-y拆开/
=(x+y)(x-4y+3)+(x-4y)+3 /将(x+y)(x-4y)+3(x+y)提取(x+y)/
=(x+y+1)(x-4y+3) /(x-4y+3)又是一个同类项，将它提取/

追问

谢啦

Answer 2

x^2-3xy-4y^2+4x-y+3
=(x+ay+b)(x+cy+d)
=x^2+(a+c)xy+acy^2+(b+d)x+(bc+ad)y+bd

所以
a+c=-3
ac=-4
a=-4,c=1或a=1,c=-4
b+d=4
bd=3
b=1,d=3或b=3,d=1
检验bc+ad=-1
可知a=-4,c=1,b=3,d=1或a=1,c=-4，b=1,d=3
所以
x^2-3xy-4y^2+4x-y+3
=(x-4y+3)(x+y+1)

Answer 3

因为x^2-3xy-4y^2=(x-4y)(x+y)
所以设原式=(x-4y+A)(x+y+B)

=x^2-3xy-4y^2+(A+B)x+(A-4B)y+AB
=x^2-3xy-4y^2+4x-y+3
所以A+B=4 A-4B=-1 AB=3
A=3 B=1
所以原式=(x-4y+3)(x+y+1)