Computing Bézier curve of a high order I have a set of ten points that much be used to compute a Bézier curve.  As you are probably aware, computing a bezier curve of order 9 is a very strenuous activity.  I need it in polynomial form.  I found this section in the wikipedia article on Bezier curves, but it cautioned against using this polynomial formula for higher order curves:
\begin{align}
\begin{split}
B(t) =& \sum \limits_{j=0}^{n}t^{j}C_{j} \\
\text{where} \\
C_j =& \frac{n!}{(n - j)!} \sum \limits_{i=0}^{j} \frac{(-1)^{i+j}P_{i}}{i!(j - i)!}
\end{split}
\end{align}

This could be practical if $C_j$ can be computed prior to many evaluations of $B(t)$; however one should use caution as high order curves may lack numeric stability (de Casteljau's algorithm should be used if this occurs).

My question is therefore, what is the standard way to compute a higher order bezier curve in polynomial form and if so, how?

I found the method to construct a set of Bezier curves in my numerical methods textbook, Numerical Analysis by Burden, Faires, and Burden (10E).
I split the points to create three connected cubic Bezier curves $C_0$, $C_1$, and $C_2$ in parametric form, where $C_i$ is represented by
\begin{align}
(x_{i}(t), y_{i}(t)) &= (a_{0}^{(i)} + a_{1}^{(i)}t + a_{2}^{(i)}t^{2} + a_{3}^{(i)}t^{3}, b_{0}^{(i)} + b_{1}^{(i)}t + b_{2}^{(i)}t^{2} + b_{3}^{(i)}t^{3})
\end{align}
for $0 \leq t \leq 1$, as determined by the left endpoint $(x_i, y_i)$, left guidepoint $(x_{i}^{+}, y_{i}^{+})$, right endpoint $(x_{i+1}, y_{i+1})$, and right guidepoint $(x_{i+1}^{-}, y_{i+1}^{-})$ for each $i = 0, 1, 2$.
\begin{align}
a_{0}^{(i)} &= x_i \\
a_{1}^{(i)} &= 3(x_{i}^{+} - x_i) \\
a_{2}^{(i)} &= 3(x_i + x_{i+1}^{-} - 2x_{i}^{+}) \\
a_{3}^{(i)} &=x_{i+1} - x_i + 3x_{i}^{+} - 3x_{i+1}^{-}
\end{align}
