\(\mathrm{f(x) = \frac{(x^2 - k^2)(x - p)}{(x - k)}}\) In the function f defined above, k and p are unique...

1. TRANSLATE the problem information

• Given: \(\mathrm{f(x) = \frac{(x^2 - k^2)(x - p)}{(x - k)}}\) where k and p are unique positive integers
• Find: Number of distinct x-intercepts

2. INFER the approach for rational functions

• For rational functions, x-intercepts occur where \(\mathrm{f(x) = 0}\)
• This means: \(\mathrm{numerator = 0}\) AND \(\mathrm{denominator \neq 0}\)
• Strategy: Find zeros of numerator, then check if any create domain issues

3. SIMPLIFY to find potential x-intercepts

• Set numerator equal to zero: \(\mathrm{(x^2 - k^2)(x - p) = 0}\)
• Factor difference of squares: \(\mathrm{x^2 - k^2 = (x - k)(x + k)}\)
• Equation becomes: \(\mathrm{(x - k)(x + k)(x - p) = 0}\)
• By zero product property: \(\mathrm{x = k}\), \(\mathrm{x = -k}\), or \(\mathrm{x = p}\)

4. INFER domain restrictions and holes

• The denominator \(\mathrm{(x - k)}\) equals zero when \(\mathrm{x = k}\)
• Since \(\mathrm{x = k}\) makes both numerator AND denominator zero, this creates a hole, not an x-intercept
• The graph has a removable discontinuity at \(\mathrm{x = k}\)

5. APPLY CONSTRAINTS to count actual x-intercepts

• Remove the hole at \(\mathrm{x = k}\)
• Remaining x-intercepts: \(\mathrm{x = -k}\) and \(\mathrm{x = p}\)
• Since k and p are unique positive integers: \(\mathrm{-k \lt 0}\) and \(\mathrm{p \gt 0}\), so these are distinct

Answer: B (2)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students find the three zeros of the numerator \(\mathrm{(k, -k, p)}\) but fail to recognize that \(\mathrm{x = k}\) creates a hole rather than an x-intercept. They don't check what happens when the denominator also equals zero at the same point.

This leads them to count all three zeros as x-intercepts and select Choice C (3).

Second Most Common Error:

Missing conceptual knowledge about rational functions: Students may not remember that x-intercepts of rational functions require the denominator to be non-zero. They might just solve the numerator equation without considering domain restrictions at all.

This also leads them to select Choice C (3).

The Bottom Line:

This problem tests the crucial distinction between zeros of the numerator and actual x-intercepts of a rational function. Success requires understanding that holes (removable discontinuities) occur when both numerator and denominator equal zero simultaneously.