A hopefully easy question. I have this:
$$c^2+(x^2+y^2+z^2)(v_x^2+v_y^2+v_z^2)-2c(x*v_x+y*v_y+z*v_z)$$
And I was wondering if this somehow factors into $$(c-f(\vec{r}, \vec{v}))^2 $$ where
$$f(\vec{r}, \vec{v})=x*v_x+y*v_y+z*v_z=\vec{r} \cdot \vec{v} $$ $$f(\vec{r}, \vec{v})^2=(x^2+y^2+z^2)(v_x^2+v_y^2+v_z^2)=(\vec{r} \cdot \vec{r})(\vec{v} \cdot \vec{v}) $$
Does such a function exist? Can such a function exist? Or am I stuck with this being the most compact form: $ c^2 + (\vec{r} \cdot \vec{r})(\vec{v} \cdot \vec{v})-2c(\vec{r} \cdot \vec{v}) $ or $ c^2 + r^2v^2-2c(\vec{r} \cdot \vec{v}) $
Expand the quadratic in powers of $c$ and equate coefficients $$(c-f(\vec{r}, \vec{v}))^2 =c^2-2cf(\vec{r}, \vec{v})+\left(f(\vec{r}, \vec{v}) \right)^2$$
$$=c^2 -2c(\vec r \cdot \vec v)+(\vec r \cdot \vec r)(\vec v \cdot \vec v)$$
this is only possible if $$(\vec r \cdot \vec v)^2=(\vec r \cdot \vec r)(\vec v \cdot \vec v)$$
Which only happens when the vectors $\vec r$ and $\vec v$ are either parallel or antiparallel, in which case $f(\vec{r}, \vec{v})=\vec{r} \cdot \vec{v}$