The expression 1/3x^2 - 2 can be rewritten as \(\frac{1}{3}(\mathrm{x} - \mathrm{k})(\mathrm{x} + \mathrm{k})\), where k is a positive constant....

1. TRANSLATE the problem information

• Given: Expression \(\frac{1}{3}x^2 - 2\)
• Need: Rewrite as \(\frac{1}{3}(x - k)(x + k)\) where \(k \gt 0\)
• Find: The value of \(k\)

2. INFER the factoring strategy

• The target form \(\frac{1}{3}(x - k)(x + k)\) shows that \(\frac{1}{3}\) is factored out
• Inside the parentheses, we need \((x - k)(x + k) = x^2 - k^2\)
• This means we need to factor out \(\frac{1}{3}\) first, then work with what remains

3. SIMPLIFY by factoring out the coefficient

• \(\frac{1}{3}x^2 - 2 = \frac{1}{3}(x^2 - 6)\)
• We divided the constant term \(-2\) by the coefficient \(\frac{1}{3}\): \(-2 \div \frac{1}{3} = -2 \times 3 = -6\)

4. INFER that we have a difference of squares

• We need to factor \(x^2 - 6\)
• This has the form \(a^2 - b^2\) where \(a = x\) and \(b^2 = 6\)
• So \(b = \sqrt{6}\)

5. SIMPLIFY using the difference of squares formula

• \(x^2 - 6 = x^2 - (\sqrt{6})^2 = (x - \sqrt{6})(x + \sqrt{6})\)
• Therefore: \(\frac{1}{3}x^2 - 2 = \frac{1}{3}(x - \sqrt{6})(x + \sqrt{6})\)

6. INFER the final answer

• Comparing \(\frac{1}{3}(x - \sqrt{6})(x + \sqrt{6})\) with \(\frac{1}{3}(x - k)(x + k)\)
• We see that \(k = \sqrt{6}\)

Answer: D. \(\sqrt{6}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students attempt to factor \(\frac{1}{3}x^2 - 2\) directly as a difference of squares without first factoring out the coefficient \(\frac{1}{3}\).

They might try: \(\frac{1}{3}x^2 - 2 = (\frac{1}{\sqrt{3}}x)^2 - (\sqrt{2})^2\) and get confused trying to make this fit the required form. This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly factor out \(\frac{1}{3}\) to get \(\frac{1}{3}(x^2 - 6)\) but then make an error with the difference of squares.

They might incorrectly think \(x^2 - 6 = (x - 6)(x + 6)\) or struggle with \(\sqrt{6}\), potentially selecting Choice A (2) by confusing it with the constant term \(-2\), or Choice B (6) by focusing on the 6 inside the parentheses.

The Bottom Line:

This problem requires students to see the two-step process: factor out the coefficient first, then apply difference of squares. The key insight is recognizing that the target form tells us exactly how to approach the factoring.