如果多项式kx²-2xy+3y²+3x-5y+2能分解成两个一次因式乘积,则k=?

kx²-2xy+3y²+3x-5y+2
=kx^2+(3-2y)x+(3y-2)(y-1)
所以,设=kx²-2xy+3y²+3x-5y+2
=(ax+(3y-2))(bx+(y-1))
=abx^2+[b(3y-2)+a(y-1)]x+(3y-2)(y-1)
=abx^2+[(3b+a)y-(2b+a)]x+(3y-2)(y-1)
所以：
ab=k
3b+a=-2
-(2b+a)=3
解方程组得：
a=-5,b=1
k=-5