P(x) = x³ +2x² -x -2الجذور الممكنة = {1, -1, 2, -2}* P(1) = 0 → x1 = 1* P(-1) = 0 → x2 = -1* P(2) = 12 → 2 ليس جذرا لكثيرة الحدودوباقي (x³ +2x² -x -2)/(x-2) هو 12* P(-2) = 0 → x3 = -2P(x) = x³ +2x² -x -2الجذور الممكنة = {1, -1, 2, -2}* P(1) = 0 → x1 = 1→ x³ +2x² -x -2 = (x-1)(x² +3x +2)هنا, r1=-1 وQ(x) = x² +3x +2* Q(-1) = 0 → x2 = -1→ x³ +2x² -x -2 = (x-1)(x² +3x +2) = (x-1)(x+1)(x+2)Having used the ""p"/"q"" result above (or, to be fair, any other means) to find all the real rational roots of a particular polynomial, it is but a trivial step further to partially factor that polynomial using those roots.
Examples:=====Finding roots without applying Ruffini's Rule=====P(x) = x³ +2x² -x -2Possible roots = {1, -1, 2, -2}*P(1) = 0 → x1 = 1*P(-1) = 0 → x2 = -1*P(2) = 12 → 2 is not a root of the polynomialand the remainder of (x³ +2x² -x -2)/(x-2) is 12*P(-2) = 0 → x3 = -2=====Finding roots applying Ruffini's Rule and obtaining a (complete) factorization=====P(x) = x³ +2x² -x -2Possible roots = {1, -1, 2, -2}*P(1) = 0 → x1 = 1Then, applying Ruffini's Rule:(x³ +2x² -x -2) / (x -1) = (x² +3x +2) →→ x³ +2x² -x -2 = (x-1)(x² +3x +2)Here, r1=-1 and Q(x) = x² +3x +2*Q(-1) = 0 → x2 = -1Again, applying Ruffini's Rule:(x² +3x +2) / (x +1) = (x +2) →→ x³ +2x² -x -2 = (x-1)(x² +3x +2) = (x-1)(x+1)(x+2)As it was possible to completely factorize the polynomial, it's clear that the last root is -2 (the previous procedure would have given the same result, with a final quotient of 1).