Which of the following is equivalent to the expression x^4 - x^2 - 6?

1. INFER the structure of the expression

• Given expression: \(\mathrm{x^4 - x^2 - 6}\)
• Key insight: This looks like a quadratic expression if we think of \(\mathrm{x^2}\) as a single unit
• Let's substitute \(\mathrm{u = x^2}\) to see this more clearly:
  • \(\mathrm{x^4 = (x^2)^2 = u^2}\)
  • \(\mathrm{x^2 = u}\)
  • So our expression becomes: \(\mathrm{u^2 - u - 6}\)

2. SIMPLIFY by factoring the quadratic

• Now we need to factor \(\mathrm{u^2 - u - 6}\)
• We're looking for two numbers that:
  • Multiply to give -6 (the constant term)
  • Add to give -1 (the coefficient of u)
• Testing factor pairs of -6:
  • 1 and -6: multiply to -6, but add to -5 ✗
  • -1 and 6: multiply to -6, but add to 5 ✗
  • 2 and -3: multiply to -6, and add to -1 ✓
• Therefore: \(\mathrm{u^2 - u - 6 = (u + 2)(u - 3)}\)

3. TRANSLATE back to the original variable

• Substitute \(\mathrm{x^2}\) back for u:
  • \(\mathrm{(u + 2)(u - 3) = (x^2 + 2)(x^2 - 3)}\)
• So \(\mathrm{x^4 - x^2 - 6 = (x^2 + 2)(x^2 - 3)}\)

Answer: B. \(\mathrm{(x^2 + 2)(x^2 - 3)}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the quadratic structure hidden in the fourth-degree polynomial. They might attempt to factor \(\mathrm{x^4 - x^2 - 6}\) directly without the substitution strategy, getting overwhelmed by the higher powers.

This leads to confusion and random guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students recognize the quadratic structure but find the wrong factor pair of -6. For example, they might use factors like (-1, 6) which multiply to -6 but add to +5 instead of -1.

This may lead them to create an incorrect factorization and select Choice A (\(\mathrm{(x^2 + 1)(x^2 - 6)}\)) or another wrong answer.

The Bottom Line:

This problem tests whether students can see past the intimidating \(\mathrm{x^4}\) term and recognize familiar patterns. The key breakthrough is realizing that substitution can transform a seemingly complex expression into a standard quadratic factoring problem.