12t2+29t-8=(3t+8)(4t-1).
Let (at±b)(ct∓d)=act2∓adt±bct-bd=12t2+29t-8, so ac=12=(1×12), (2×6), or (3×4); and bd or db=(1×8) or (2×4).
We can write act2∓adt±bct-bd= act2±(bc-ad)t-bd or act2∓(ad-bc)t-bd.
A table shows all the possible combinations:
| a |
b |
c |
d |
|ad-bc| |
| 1 |
1 |
12 |
8 |
4 |
| 2 |
1 |
6 |
8 |
10 |
| 3 |
1 |
4 |
8 |
20 |
| 1 |
8 |
12 |
1 |
95 |
| 2 |
8 |
6 |
1 |
46 |
| 3 |
8 |
4 |
1 |
29 |
| 1 |
2 |
12 |
4 |
20 |
| 2 |
2 |
6 |
4 |
4 |
| 3 |
2 |
4 |
4 |
4 |
| 1 |
4 |
12 |
2 |
46 |
| 2 |
4 |
6 |
2 |
20 |
| 3 |
4 |
4 |
2 |
10 |
|ad-bc| simply means the positive difference between these products.
The red highlighted row is the one we want because 29 is the middle coefficient. Therefore a=3, b=8, c=4 and d=1.
ad-bc=3-32=-29 which is the correct sign for the middle term of the quadratic. So in act2∓(ad-bc)t-bd, ∓ can be replaced by + because ad-bc is already negative, and ± can be replaced by -. Therefore we have (at-b)(ct+d)=(3t-8)(4t+1) as the factorisation.