examples:

\[ex:3\] The elliptic translation of does not admit a factorization if $a \neq b$. If $a = b > 0$, it admits infinitely many factorizations $$C = t^2 + 1 + a{\varepsilon}({\mathbf{i}}+ {\mathbf{j}}t)
      = (t - {\mathbf{k}}+ {\varepsilon}(\alpha{\mathbf{i}}+ (\beta+a){\mathbf{j}}))
        (t + {\mathbf{k}}- {\varepsilon}(\alpha{\mathbf{i}}+ \beta{\mathbf{j}}))$$ with $\alpha$, $\beta \in {\mathbb{R}}$. The Viviani motion in admits only the factorization $$\label{eq:3}
    C = t^2 - ({\mathbf{j}}+ {\mathbf{k}}- {\varepsilon}({\mathbf{j}}+ {\mathbf{k}}))t - {\mathbf{i}}= (t - {\mathbf{k}}+ {\varepsilon}{\mathbf{j}})(t - {\mathbf{j}}- {\varepsilon}{\mathbf{k}}).$$ By considering primal parts only (the dual parts are merely there because the sphere center is not the affine origin), we see that it is the composition of rotations with equal angular speed about the second and the third coordinate axis.

An algorithm for computing factorizations of generic motion polynomials has been presented in [@hegedus13]. It is displayed in and . The assumptions on $M$ and $C$ in guarantee that the remainder $R$ in Line 1 has an invertible leading coefficient (compare [@hegedus13 Theorem 3]). They are met in because $C$ is assumed to be generic. The non-uniqueness of the factorization comes from the undetermined order of the quadratic factors in Line \[line:quadfac\] of . In Line \[line:lquo\] we compute the quotient of polynomial 