12a^2-13a-14=(4a-7)(3a+2)
How is this solved?
First, list the factors of 12 and 14:
(1,12), (2,6), (3,4) and (1,14), (2,7).
Play these factors off against each other, but in the case of the factors of 14, include the reverse order:
(1,12), (2,6), (3,4) and (1,14), (2,7), (14,1), (7,2).
Represent the factors of 12 as (A,B) and the factors of 14 as (C,D).
Now consider the expressions AD-BC and AD+BC. Look at the signs between the terms. The sign in front of the constant is minus (-14), which tells us that the signs in the factor brackets must be different. The sign in front of the constant also tells us to use AD-BC as the expression. Now we make a table with all the possible values of A, B, C, D:
| ROW |
A |
B |
C |
D |
AD |
BC |
|AD-BC| |
| 1 |
1 |
12 |
1 |
14 |
14 |
12 |
2 |
| 2 |
1 |
12 |
2 |
7 |
7 |
24 |
17 |
| 3 |
1 |
12 |
14 |
1 |
1 |
168 |
167 |
| 4 |
1 |
12 |
7 |
2 |
2 |
84 |
82 |
| 5 |
2 |
6 |
1 |
14 |
28 |
6 |
22 |
| 6 |
2 |
6 |
2 |
7 |
14 |
11 |
3 |
| 7 |
2 |
6 |
14 |
1 |
2 |
84 |
82 |
| 8 |
2 |
6 |
7 |
2 |
4 |
42 |
38 |
| 9 |
3 |
4 |
1 |
14 |
42 |
4 |
38 |
| 10 |
3 |
4 |
2 |
7 |
21 |
8 |
13 |
| 11 |
3 |
4 |
14 |
1 |
3 |
56 |
53 |
| 12 |
3 |
4 |
7 |
2 |
6 |
28 |
22 |
Next, we look for the row in which the last column contains the coefficient of the middle term (13). It's row 10. That tells us the required factors (Aa C)(Ba D), but not the signs. The sign on the middle term (-13a) goes in front of whichever of C or D produces the larger product of AD or BC. Since 21 is bigger than 8 in row 10, minus goes in front of D, so we have (Aa+C)(Ba-D)=(3a+2)(4a-7).