对关于x的方程
6x^2-yx+(-2y^2+my-6)=0,(1)
Delta=y^2-24(-2y^2+my-6)
=49y^2-24my+144.
则(1)的两根为
x1=(y+根号(Delta))/12,
x2=(y-根号(Delta))/12,
则 6x^2-yx+(-2y^2+my-6)=6(x-x1)(x-x2).
由于6x^2-yx+(-2y^2+my-6)能分解成两个一次因式的积,
则 x-x1,x-x2为一次因式.
所以 Delta=49y^2-24my+144为一次因式的平方.
所以 Delta2=(-24m)^2-4*49*144=0.
解得 m1=7,m2=-7.
(i)当m=7时,Delta=(7y-12)^2,此时
x1=(2/3)y-1,
x2=(-1/2)y+1.
则 6x^2-xy-2y^2+7y-6
=6(x-(2/3)y+1)(x+(1/2)y-1)
=(3x-2y+3)(2x+y-2).
(ii)当m=-7时,Delta=(7y+12)^2,此时
x1=(2/3)y+1,
x2=(-1/2)y-1.
则 6x^2-xy-2y^2-7y-6
=6(x-(2/3)y-1)(x+(1/2)y+1)
=(3x-2y-3)(2x+y+2).