问题描述：
(2x^4-3x^3+6x^2+7x-4b)÷(x-2)怎么算?
求结果,题应该没错……
求具体一点，我可看不懂……
=[2x^3(x-2)+x^3 这一步是怎么搞得？对不起我不太聪明……
问题解答：
我来补答
用长除法.
(2x^4-3x^3+6x^2+7x-4b)/(x-2)
=[2x^3(x-2)+x^3+6x^2+7x-4b]/(x-2)
=2x^3+(x^3+6x^2+7x-4b)/(x-2)
=2x^3+[x^2(x-2)+8x^2+7x-4b]/(x-2)
=2x^3+x^2+(8x^2+7x-4b)/(x-2)
=2x^3+x^2+[8x(x-2)+23x-4b]/(x-2)
=2x^3+x^2+8x+(23x-4b)/(x-2)
=2x^3+x^2+8x+[23(x-2)+46-4b]/(x-2)
=2x^3+x^2+8x+23+(46-4b)/(x-2)
你有没有写错?4b是不是46?如果是,那就等于
2x^3+x^2+8x+23
如果没错就是
2x^3+x^2+8x+23+(46-4b)/(x-2)
补充：2x^4-3x^3
=2x^4-4x^3+x^3
=2x^3(x-2)+x^3
其他的以此类推.
[2x^3(x-2)+x^3+6x^2+7x-4b]/(x-2)
=[2x^3(x-2)]/(x-2)+(x^3+6x^2+7x-4b)/(x-2)
=2x^3+(x^3+6x^2+7x-4b)/(x-2)
