or is there a utility in `sage`

that returns the bezier control points of a curve given that curve?

im writing a program that takes a curve in polynomial form and returns those control points, this is the part in question:

```
def bezierControlPoints(poly):
a = sage.var('a')
b = sage.var('b')
c = sage.var('c')
d = sage.var('d')
t = sage.var('t')
bezier = ((1 - t) ** 3) * a + 3 * ((1 - t) ** 2) * t * \
b + 3 * ((1 - t) * (t ** 2)) * c + (t ** 3) * d
eqn1 = poly.substitute(t=0)
eqn2 = poly.substitute(t=.33)
eqn3 = poly.substitute(t=.66)
eqn4 = poly.substitute(t=1)
cb1 = bezier.substitute(t=0)
cb2 = bezier.substitute(t=.33)
cb3 = bezier.substitute(t=.66)
cb4 = bezier.substitute(t=1)
solns = sage.solve([eqn1 == cb1, eqn2 == cb2, eqn3 == cb3,
eqn4 == cb4], a, b, c, d, solution_dict=True)
return [[s[a].n(30), s[b].n(30), s[c].n(30), s[d].n(30)] for s in solns]
```

but the control points are coordinates, and this returns single numbers like coefficients. for example

```
f(x) = x^4 - 10*x^3 + 35*x^2 - 50*x + 24
```

returns

```
[24.000000, 3.8069265, -4.2949254, -0.30555556]
[-0.30555556, 0.85185185, 0.85185185, -0.30555556]
[-0.30555556, -1.3680046, -1.2661527, 0.00000000]
```

so how do i finish processing these portions of the control points?

**background**

the parts of the program i have written that i didnt share take the polynomial of arbitrary order, break it up into simpler subcurves, parameterize those subcurves, and those parameterized subcurves are the arguements of the `bezierControlPoints()`

, which is called iteratively depending on the order of the polynomial.