Editing Rational root theorem (section)

=== Proof using Gauss's lemma ===

Should there be a nontrivial factor dividing all the coefficients of the polynomial, then one can divide by the [[greatest common divisor]] of the coefficients so as to obtain a primitive polynomial in the sense of [[Gauss's lemma (polynomial)|Gauss's lemma]]; this does not alter the set of rational roots and only strengthens the divisibility conditions. That lemma says that if the polynomial factors in {{math|'''Q'''[''X'']}}, then it also factors in {{math|'''Z'''[''X'']}} as a product of primitive polynomials. Now any rational root {{math|''p''/''q''}} corresponds to a factor of degree 1 in {{math|'''Q'''[''X'']}} of the polynomial, and its primitive representative is then {{math|''qx'' − ''p''}}, assuming that {{math|''p''}} and {{math|''q''}} are coprime. But any multiple in {{math|'''Z'''[''X'']}} of {{math|''qx'' − ''p''}} has leading term divisible by {{math|''q''}} and constant term divisible by {{math|''p''}}, which proves the statement. This argument shows that more generally, any irreducible factor of {{math|''P''}} can be supposed to have integer coefficients, and leading and constant coefficients dividing the corresponding coefficients of&nbsp;{{math|''P''}}.