用牛顿迭代法求一元三次方程的根要求：1.用牛顿迭代法求方程ax³+bx²+cx+d=0的根.、2.求方程系数a=1,b=

用牛顿迭代法求方程'a * x ^ 3 + b * x ^ 2 + c * x + d = 0,系数a = 1,b = 2,c = 3,d = 4,x在0附近的一个实数根为1.33333333333.算法代码如下：
Private Sub Command1_Click() '牛顿迭代法
Dim a As Double,b As Double,c As Double,d As Double,xx1 As Double
Dim n As Long
a = 1
b = 2
c = 3
d = 4
xx1 = 0 '初始值为0,x在0附近
Print Nim(a,b,c,d,xx1,n) '方程在XX1附近的根
Print n '迭代次数
End Sub
'牛顿迭代法Newton iteration method
Function Nim(ByVal a As Double,ByVal b As Double,ByVal c As Double,ByVal d As Double,ByVal x0 As Double,ByRef n As Long) As Double
Dim x As Double,y As Double,dy As Double,ydy As Double,i As Long
n = -1 '迭代次数
x = x0
For i = 0 To 1000
y = a * x ^ 3 + b * x ^ 2 + c * x + d
dy = 3 * a * x ^ 2 + 2 * b * x + c
If Abs(x - x0) 1000 Then n = -2 '该方程无解解
End Function