Always divisible? — For natural number x, analyze the expression 6x^2 + 6x and determine which fixed divisibility statement is always true.

Introduction / Context:
Divisibility questions often become straightforward after factoring. Recognizing common factors and parity properties lets us determine strong, universal divisibility claims for expressions that depend on natural numbers. Here we analyze 6x^2 + 6x and decide which divisors always occur.

Given Data / Assumptions:

• x is a natural number.
• Expression: 6x^2 + 6x.
• We must pick the statement that is always true for all x.

Concept / Approach:
Factor the expression first, then use the property of consecutive integers. The product x(x + 1) involves two consecutive integers, which guarantees evenness (and hence at least one factor 2). Combine this with the 6 outside to determine the minimal universal power-of-2 and power-of-3 factors, and thus the largest guaranteed composite divisor.

Step-by-Step Solution:
Factor: 6x^2 + 6x = 6x(x + 1).Among consecutive integers x and x+1, one is even → x(x+1) is even → contributes at least one factor 2.Therefore 6x(x+1) has factors 6 * 2 = 12 at minimum, for any x.Hence the expression is always divisible by 12. Divisible by 12 implies divisible by 6 as well.

Verification / Alternative check:
Test x=1: 6*1^2 + 6*1 = 12 (divisible by 12). Test x=2: 24 + 12 = 36 (divisible by 12). Test x=3: 54 + 18 = 72 (divisible by 12). These checks match the general proof.

Why Other Options Are Wrong:

• 12 only: Misleading wording—while the expression is always divisible by 12, it is also divisible by 6; “only” suggests otherwise.
• 6 only / 3 only: Too weak; a stronger always-true statement is that it is divisible by 12.
• 24: Not guaranteed (counterexample x=1 gives 12).

Common Pitfalls:
Stopping after extracting factor 6; forgetting the evenness of x(x+1); over-claiming divisibility by 24 without verifying small x.

Final Answer:
6 and 12