Well, the "eigenvalue equation" for an n by n matrix is an nth degree polynomial equation. It is far easier to solve a quadratic equation than a cubic equation! That was the only reason, I suspect, to starting with one of the eigenvalues "given". Dividing a cubic equation by "x- a", where a is a solution, reduces it to a quadratic equation for the other two solutions.
I'm not clear what you mean by "free variables" here. That term is used in a number of different ways in different mathematical problems. It is true that if a given eigenvalue is a "double" root of the eigenvalue equation, then the space of eigenvectors corresponding to that eigenvalue may be of dimension 1 or 2. I suspect that it is in the case that a double root has eigenspace of dimension only 1 that your "free variables" arise.
As a rather trivial example, the matrix has "2" as a double eigenvalue and both <1, 0, 0> and <0, 1, 0> are independent eigenvectors.
On the other hand, also has "2" as a double eigenvalue but now the space of eigenvectors corresponding to eigenvalue 2 is only one dimensional and is spanned by <1, 0, 0>.