Factoring by the trial and-error method

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