Rational Root Theorem

This is the systematic version of the guessing in §3.8: instead of testing arbitrary numbers, list every divisor of the constant term $a_0$ as a candidate numerator, every divisor of the leading coefficient $a_n$ as a candidate denominator, form every possible fraction $p/q$, and test only those.

Special case: monic equations. If $a_n=1$, the theorem forces $q=\pm1$ for any rational root $p/q$, so $p/q$ must itself be an integer, and that integer must divide $a_0$. So a monic integer-coefficient polynomial can never have a non-integer rational root — any rational root is a whole-number divisor of the constant term.

Worked illustration. For $x^2-5x-6=0$, divisors of $6$ are $\pm1,\pm2,\pm3,\pm6$ — the only candidates the theorem allows. Testing shows $-1$ and $6$ actually work; the theorem narrows the search, it does not guarantee every candidate is a root (nor even that any candidate is — e.g. for $x^2+4=0$ the candidates $\pm1,\pm2,\pm4$ include no actual root at all, since both roots are imaginary). …