Factoring by the trial and-error method

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Factoring by the trial and-error method

Factoring by the trial and-error method

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  • 1. Factoring by the Trial-and-error method TE Factorization Formula Let f = ax2+bx+c be a quadratic trinomial with integer coefficients. If there exist integers d, e, m and n satisfying the three conditions de = a mn = c dn+me = b then the trinomial f can be factored over the integers as ax2+bx+c = (dx+m)(ex+n) Algorithm TE Given a quadratic trinomial with integer coefficients f = ax2+bx+c this algorithm will find a factorization of f over the integers if one exists, else it will determine that f is not factorable over the integers. Step 1. Find the prime factorization of |a|, the absolute value of the quadratic coefficient of f. Then, taking into account the sign of a, list all factorizations a = de with 0 < d ≤ |e| of a into the product of two integers d and e for which d is positive and is less than or equal to the absolute value of e. Step 2. Find the prime factorization of |c|, the absolute value of the constant term of f. Then, taking into account the sign of c, list all factorizations c = mn with m>0 of c into the product of two integers m and n for which m is positive. Step 3. For each factorization a = de in Step 1 and each factorization c = mn in Step 2, • Compute the cross-product sum s = dn+me. • If s = b, terminate with the factorization ax2+bx+c = (dx+m)(ex+n) • If s = –b, terminate with the factorization ax2+bx+c = (dx–m)(ex–n) Step 4. If s ≠ b and s ≠ –b for all factorizations a = de (in Step 1) and all factorizations c = mn (in Step 2), then the trinomial f is nonfactorable over the integers. Quick Reference on Factoring Trinomials • © 2001 K-14 Publishing Company
  • 2. 2 f = 14x2+11x–15 with a = 14, b = 11, c = –15 Example Step 1. Prime factorization of |a| = |14| = 14: 14 = 2(7) Factorizations a = de of a = 14 for which 0 < d ≤ |e|: 14 = 1(14) = 2(7) Step 2. Prime factorization of |c| = |–15| = 15: 15 = 3(5) Factorizations c = mn of c = –15 for which m > 0: –15 = 1(–15) = 15(–1) = 3(–5) = 5(–3) Step 3. For each factorization a = de in Step 1 and each factorization c = mn in Step 2, compute the cross-product sum s = dn+me. (d,e) (m,n) (1,14) (2,7) (1,–15) 1(–15) + 1(14) = –1 2(–15) + 1(7) = –23 (15,–1) 1(–1) + 15(14) = 209 2(–1) + 15(7) = 103 (3,–5) 1(–5) + 3(14) = 37 2(–5) + 3(7) = 11 (5,–3) 1(–3) + 5(14) = 67 2(–3) + 5(7) = 29 Table 1. Cross-product sums s = dn+me for f = 14x2+11x–15 From Table 1, we find that the factorization a = 14 = 2(7) with d = 2 and e = 7, and the factorization c = –15 = 3(–5) with m = 3 and n = –5, result in the cross-product sum s = dn+me = 2(–5)+3(7) = –10+21 = 11 = b Factorization for f: 14x2+11x–15 = (2x+3)(7x–5) Quick Reference on Factoring Trinomials • © 2001 K-14 Publishing Company