Vous avez cherché: factorize (Anglais - Arabe)

factorize
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factorize, factor
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in this exercise you have to factorize a given number.
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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).
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.