Question

初一的因式分解题，急急急！！！

x²-xy-6y²+2x+14y-8

可以用完全平方，平方差和十字相乘法

Answer 1

利用双十字相乘法分解因式.将上式按x降幂排列，并把y当作常数，可变形为
x^2+(2-y)x-(6y^2-14y+8)
=x^2+(2-y)x-(2y-2)(3y-4).从中可发觉到(2y-2)-(3y-4)正好等于x前面的系数(2-y),
所以,原式等于[x-(3y-4)][x+(2y-2)]
=(x-3y+4)(x+2y-2)

Answer 2

x²-xy-6y²+2x+14y-8
=(x-3y)(x+2y)+2x+14y-8
=(x-3y)(x+2y)+4x+8y-2x+6y-8
=(x-3y)(x+2y)+4(x+2y)-2(x-3y+4)
=(x+2y)(x-3y+4)-2(x-3y+4)
=(x+2y-2)(x-3y+4)

Answer 3

x²-xy-6y²+2x+14y-8
=(x+2y)(x-3y)+2x+14y-8
=(x+2y-2)(x-3y+4)