Let's factor the equation \( 12b^2 = 24 + 15 \).

### Step 1: Simplify the Right-Hand Side
First, simplify the equation by adding the constants on the right-hand side:
\[
12b^2 = 24 + 15
\]
\[
12b^2 = 39
\]

### Step 2: Bring All Terms to One Side
Subtract 39 from both sides to set the equation to zero:
\[
12b^2 - 39 = 0
\]

### Step 3: Factor the Equation
To factor the equation, notice that both terms share a common factor of 3:
\[
3(4b^2 - 13) = 0
\]

The equation is now factored as \( 3(4b^2 - 13) = 0 \).

### Conclusion:
The factorization of the equation \( 12b^2 = 24 + 15 \) is \( 3(4b^2 - 13) = 0 \). The equation has been fully factored over the integers.

Would you like more details or have further questions?

### Related Questions:
1. How do you solve \( 3(4b^2 - 13) = 0 \) for \( b \)?
2. Can the quadratic term \( 4b^2 - 13 \) be factored further?
3. How does the distributive property apply when factoring expressions?
4. What are the implications if \( b^2 \) had a negative coefficient?
5. How do you check if a factorization is correct?

### Tip:
Always check for a common factor before proceeding with more complex factoring methods!

Learn how to factor the equation 12b^2 = 24 + 15 with a step-by-step solution. This problem covers algebraic concepts such as factoring quadratic equations. Suitable for students in Grades 7-9.

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