因式分解：x²-2xy-3y²+2x+10y-8

Question

RT
要过程
快

Answer 1

x^2-2xy-3y^2+2x+10y-8
=x^2+2(1-y)x-3y^2+10y-8
=x^2+2(1-y)x+(1-y)^2-(1-y)^2-3y^2+10y-8
=[x+(1-y)]^2-4y^2+12y-9
=[x+(1-y)]^2-(2y-3)^2
=[x+(1-y)-(2y-3)]*[x+(1-y)+(2y-3)]
=(X-3Y+4)(X+Y-2)
^表示次方
^2表示平方

Answer 2

x²-2xy-3y²+2x+10y-8
因为x²-2xy-3y²=（x-3y)(x+y)
所以x²-2xy-3y²+2x+10y-8=（x-3y+a)(x+y+b).........待定系数法
（x-3y+a)(x+y+b）=x²-2xy-3y²+（a+b)x+(a-3b)y+ab
a+b=2
a-3b=10
ab=-8
a=4,b=-2
x²-2xy-3y²+2x+10y-8=（x-3y+4)(x+y-2)

Answer 3

先分解x²-2xy-3y²
x²-2xy-3y²=（x-3y)（x+y)
再分解原式=（x-3y+4)（x+y-2) 这一步要用到双十字分解或待定系数

Answer 4

x²-2xy-3y²+2x+10y-8
=(x-3y)(x+y)+2x+10y-8
=(x-3y+4)(x+y-2)