Let \(\mathrm{h(x) = x^4 - 1}\) and \(\mathrm{k(x) = x^2 - 2x + 1}\). For x gt 1, which of...

1. INFER the solution strategy

• Given: \(\mathrm{h(x) = x^4 - 1}\) and \(\mathrm{k(x) = x^2 - 2x + 1}\)
• Need to find: \(\mathrm{h(x)/k(x)}\) for \(\mathrm{x \gt 1}\)
• Key insight: Both polynomials can be factored, which will allow cancellation in the rational expression

2. SIMPLIFY the numerator \(\mathrm{h(x) = x^4 - 1}\)

• Recognize this as a difference of squares: \(\mathrm{x^4 - 1 = (x^2)^2 - 1^2}\)
• Apply difference of squares: \(\mathrm{(x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1)}\)
• Factor \(\mathrm{x^2 - 1}\) further (also difference of squares): \(\mathrm{x^2 - 1 = (x - 1)(x + 1)}\)
• Complete factorization: \(\mathrm{h(x) = (x - 1)(x + 1)(x^2 + 1)}\)

3. SIMPLIFY the denominator \(\mathrm{k(x) = x^2 - 2x + 1}\)

• Recognize this as a perfect square trinomial
• Pattern check: first term = \(\mathrm{x^2}\), middle term = \(\mathrm{-2x}\), last term = \(\mathrm{1}\)
• This fits \(\mathrm{a^2 - 2ab + b^2}\) where \(\mathrm{a = x}\) and \(\mathrm{b = 1}\)
• Therefore: \(\mathrm{k(x) = (x - 1)^2}\)

4. SIMPLIFY the rational expression

• Set up the division: \(\mathrm{h(x)/k(x) = \frac{(x - 1)(x + 1)(x^2 + 1)}{(x - 1)^2}}\)
• Cancel one \(\mathrm{(x - 1)}\) factor from both numerator and denominator
• Result: \(\mathrm{\frac{(x + 1)(x^2 + 1)}{(x - 1)}}\)

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students don't recognize that \(\mathrm{x^4 - 1}\) requires factoring in two stages (difference of squares applied twice). They might try to factor it as \(\mathrm{(x^2 - 1)(x^2 + 1)}\) but then stop without factoring \(\mathrm{x^2 - 1}\) further.

This incomplete factoring means they can't properly cancel with the denominator, leading them to select Choice A: \(\mathrm{\frac{(x^2 + 1)}{(x - 1)}}\) which represents their incomplete simplification.

Second Most Common Error:

Missing conceptual knowledge: Students don't recognize \(\mathrm{x^2 - 2x + 1}\) as a perfect square trinomial and may try to factor it incorrectly or leave it unfactored.

This prevents proper cancellation and may lead them to select Choice C: \(\mathrm{\frac{(x + 1)}{(x - 1)}}\) if they somehow manage partial cancellation through algebraic manipulation mistakes.

The Bottom Line:

This problem tests pattern recognition for special polynomial forms. Success depends on systematically applying factoring patterns and being thorough with multi-step factorizations rather than stopping at the first factoring opportunity.