Question

Given the polynomial expression 6a2 + 6ca − 12a − 12c, factor completely.

6a(a – 2)(a + c)
6a(a – 2)(a – c)
6(a – 2)(a – c)
6(a – 2)(a + c)

Answer 1

To factor the polynomial expression \( 6a^2 + 6ca - 12a - 12c \) completely, we can start by rearranging and grouping the terms:

1. Start with the original expression:
\[
6a^2 + 6ca - 12a - 12c
\]

2. We can group the terms:
\[
(6a^2 - 12a) + (6ca - 12c)
\]

3. Factor out the common factors in each group:
\[
6a(a - 2) + 6c(a - 2)
\]

4. Next, we notice that \( (a - 2) \) is a common factor:
\[
6(a - 2)(a + c)
\]

Thus, the completely factored form of the polynomial is:
\[
6(a - 2)(a + c)
\]

Considering the options provided:
- \( 6(a - 2)(a + c) \) is the correct factored form.

So, the answer is:
\[
\boxed{6(a - 2)(a + c)}
\]