# Generalizing the de Casteljau algorithm to cubic Bézier curve trisection

Splitting the Bézier cubic defined by the four control points $$P, Q, R$$, and $$S$$ in two parts corresponding to the two parametric subintervals $$[0, t]$$ and $$[t, 1]$$ is relatively easy: we perform linear interpolations that result in new control points $$P, PQ, PQR, PQRS$$ and $$PQRS, QRS, RS, S$$.

Now I want to generalize to the case of the three subintervals $$[0,t], [t, u]$$ and $$[u,1]$$ in a single go. The control points for the left and right arcs are found as above, using the fractions $$t$$ and $$u$$, i.e. $$P,PQ_t,PQR_t,PQRS_t$$ and $$PQRS_u,QRS_u, RS_u, S$$. What I am missing is how to compute the two remaining control points for the middle arc.

I know that I could split as $$[0,t]$$ and $$[t,1]$$, then split the second subinterval using $$[0,\frac u{1-t}]$$ and $$[\frac u{1-t}]$$, but I think/hope that a direct approach is possible.