I need to solve $p(x)=aq(x)$ with multiple real $a$, where $p(x)$ and $q(x)$ are the two polynomials in $x$ (with real coefficients). The roots of $p(x)$ and $q(x)$ were found previously, i.e. these two polynomials are factored.
Is it somehow possible to solve the equation in question without multiplying $q(x)$ by $a$ and transferring everything to the left for every new $a$? Is there any workaround?
It's quite frustrating that the two polynomials $p(x)$ and $q(x)$ are known (together with their roots), but every time I substitute a new parameter value ($a$ in this case) I need to solve a totally new polynomial.
Sorry if the question is stupid. Actually I don't really believe that there's any workaround, although finding one would be really nice.
The problem is solved numerically, all polynomial roots are found as eigenvalues of the polynomial's companion matrix.
Sadly no. If there was such a shortcut, solving any polynomial equation would be trivial. For particular forms of your polynomials it might be possible, but not in general.