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

Example f = 14x2+11x�15 with a = 14, b = 11, c = �15

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