When applying the rule of signs to a polynomial, one can determine possible positive and negative roots.
But we can also apply a substitution or index shifting as follows:$$P(x)\to P(x\pm n)$$
Then apply the rule of signs again, and again, and again, until we've narrowed our possible roots down to small regions of the graph.
So my question is whether or not this could work for isolating roots and if it is effective.
It can work. Consider $f(x)=(x-2)(x+2)(x-4)=x^3-4x^2-4x+16$ The law of signs says it has $0$ or $2$ positive roots. $f(-x)=-x^3-4x^2+4x+16$, so there is exactly $1$ negative root. Now let $y=x-3$. We have $f(y)=(y+1)(y+5)(y-1)=y^3+5y^2-y-5$ It has exactly one positive root, so the original has one root with $x \gt 3$ Now we know the original has one root with $x \lt 0$, one root with $x \in (0,3)$ and one root with $x \gt 3$. The shift has given us new information.
Whether it is effective depends on how clever you are choosing the shifts. If we made a large shift in either direction, we would learn there were $1$ or $3$ roots on the side with the original origin and none on the other side, which would not help much.