The expression $12y^2-65y+42$ can be written as $(Ay-14)(By-3),$ where $A$ and $B$ are integers. What is $AB + A$?
We see that $12y^2-65y+42=(3y-14)(4y-3)$, thus $A = 3$ and $B = 4$. Hence, $AB + A = \boxed{15}.$
15
Level 3
Algebra
Problem: The expression $12y^2-65y+42$ can be written as $(Ay-14)(By-3),$ where $A$ and $B$ are integers. What is $AB + A$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for integers A and B such that the expression $12y^2 - 65y + 42$ can be factored into the form $(Ay - 14)(By - 3)$, we first expand the right-hand side:

1. Expand $(Ay - 14)(By - 3)$:
   - This gives $AB y^2 - (14B + 3A)y + 42$.

2. Set the coefficients equal to those in the original expression:
   - For the $y^2$ term: $AB = 12$.
   - For the $y$ term: $-(14B + 3A) = -65$, which simplifies to $14B + 3A = 65$.
   - For the constant term: $42$ matches.

3. We now have a system of equations:
   - Equation 1: $AB = 12$.
   - Equation 2: $14B + 3A = 65$.

4. Solve for integer pairs (A, B) that satisfy Equation 1:
   - Possible pairs are: (1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12, 1).

5. Test each pair in Equation 2:
   - For (1, 12): $14(12) + 3(1) = 168 + 3 = 171$ (not valid).
   - For (2, 6): $14(6) + 3(2) = 84 + 6 = 90$ (not valid).
   - For (3, 4): $14(4) + 3(3) = 56 + 9 = 65$ (valid).
   - For (4, 3): $14(3) + 3(4) = 42 + 12 = 54$ (not valid).
   - For (6, 2): $14(2) + 3(6) = 28 + 18 = 46$ (not valid).
   - For (12, 1): $14(1) + 3(12) = 14 + 36 = 50$ (not valid).

6. The valid pair is (3, 4). Thus, A = 3 and B = 4.

7. Calculate AB + A:
   - $AB = 3 \times 4 = 12$.
   - $AB + A = 12 + 3 = 15$.

The final answer is 15.<|END_OF_LATENT|> The final answer is $15$.