roots and only strengthens the divisibility conditions. That lemma says that if the polynomial factors in , then it also factors in as a product of primitive polynomials. Now any rational root corresponds to a factor of degree 1 in of the polynomial, and its primitive representative is then , assuming that and are coprime. But any multiple in of has leading term divisible by and constant term divisible by , which proves the statement. This argument shows that more generally, any irreducible factor of can be supposed to have integer coefficients, and leading and constant coefficients dividing the corresponding coefficients of . Examples First In the polynomial any rational root fully reduced would have to have a numerator that divides evenly into 1 and a denominator that divides evenly into 2. Hence the only possible rational roots are ±1/2 and ±1; since neither of these equates the polynomial to zero, it has no rational roots. Second In the polynomial the only possible rational roots would have a numerator that divides 6 and a denominator that divides 1, limiting the possibilities to ±1, ±2, ±3, and ±6. Of these, 1, 2, and –3 equate the polynomial to zero, and hence are its rational roots. (In fact these are its only roots since a cubic has only three roots; in general, a polynomial could have some rational and some irrational roots.) Third Every rational root of the polynomial must be among the numbers These 8 root candidates can be tested by evaluating , for example using Horner's method. It turns out there is exactly one with . This process may be made more efficient: if , it can be used to shorten the list of remaining candidates. For example, does not work, as . Substituting yields a polynomial in with constant term , while the coefficient of remains the same as the coefficient of . Applying the rational root theorem thus yields the possible roots , so that True roots must occur on both lists, so list of rational root candidates has shrunk to just and . If rational roots are found, Horner's method will also yield a polynomial of degree whose roots, together with the rational roots, are exactly the roots of the original polynomial. If none of the candidates is a solution, there can be no rational solution. See also Fundamental theorem of algebra Integrally closed domain Descartes' rule