∵最高次是3
则设M=ay^2+by+c
则M(y+2)-(y-1)
=(ay^2+by+c)(y+2)-y+1
=ay^3+2ay^2+by^2+2by+cy+2c-y+1
=ay^3+(2a+b)y^2+(2b+c-1)y+2c+1
=2y^3+3y^2+7
a=2
2a+b=3
b=-1
2b+c-1=0
c=3
所以
M=2y^2-y+3
∵最高次是3
则设M=ay^2+by+c
则M(y+2)-(y-1)
=(ay^2+by+c)(y+2)-y+1
=ay^3+2ay^2+by^2+2by+cy+2c-y+1
=ay^3+(2a+b)y^2+(2b+c-1)y+2c+1
=2y^3+3y^2+7
a=2
2a+b=3
b=-1
2b+c-1=0
c=3
所以
M=2y^2-y+3