How many solutions does the equation \(\mathrm{(x-4)(2x+1) = (x-4)(2x+7)}\) have?

1. INFER the most efficient approach

• Given equation: \((\mathrm{x}-4)(2\mathrm{x}+1) = (\mathrm{x}-4)(2\mathrm{x}+7)\)
• Key insight: Both sides have the common factor \((\mathrm{x}-4)\)
• Strategy: Move everything to one side and factor out the common term, rather than expanding

2. SIMPLIFY by rearranging

• Subtract the right side from both sides:
  \((\mathrm{x}-4)(2\mathrm{x}+1) - (\mathrm{x}-4)(2\mathrm{x}+7) = 0\)

3. SIMPLIFY by factoring out the common factor

• Factor out \((\mathrm{x}-4)\):
  \((\mathrm{x}-4)[(2\mathrm{x}+1) - (2\mathrm{x}+7)] = 0\)
• Simplify the bracketed expression:
  \((\mathrm{x}-4)[2\mathrm{x} + 1 - 2\mathrm{x} - 7] = 0\)
  \((\mathrm{x}-4)[-6] = 0\)

4. INFER the solution using zero product property

• Since we have \((\mathrm{x}-4)(-6) = 0\), either:
  • \(\mathrm{x}-4 = 0\), or
  • \(-6 = 0\)
• Since \(-6 \neq 0\), we must have \(\mathrm{x}-4 = 0\)
• Therefore: \(\mathrm{x} = 4\)

5. Verify the solution

• When \(\mathrm{x} = 4\): Both sides equal \((0)(9) = 0\) and \((0)(15) = 0\) ✓

Answer: A (Exactly one)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students see the common factor \((\mathrm{x}-4)\) on both sides and incorrectly conclude that the equation is always true, thinking 'since both sides have the same factor, they're always equal.'

This misconception leads them to believe there are infinitely many solutions, causing them to select Choice (C) (Infinitely many).

Second Most Common Error:

Poor SIMPLIFY execution: Students attempt to expand both sides instead of factoring, leading to:

• Left side: \((\mathrm{x}-4)(2\mathrm{x}+1) = 2\mathrm{x}^2 + \mathrm{x} - 8\mathrm{x} - 4 = 2\mathrm{x}^2 - 7\mathrm{x} - 4\)
• Right side: \((\mathrm{x}-4)(2\mathrm{x}+7) = 2\mathrm{x}^2 + 7\mathrm{x} - 8\mathrm{x} - 28 = 2\mathrm{x}^2 - \mathrm{x} - 28\)

Setting them equal: \(2\mathrm{x}^2 - 7\mathrm{x} - 4 = 2\mathrm{x}^2 - \mathrm{x} - 28\), which simplifies to \(-6\mathrm{x} = -24\), giving \(\mathrm{x} = 4\). However, many students make arithmetic errors in this longer process and may get confused or select Choice (D) (Zero) if they think their algebra led to a contradiction.

The Bottom Line:

This problem rewards recognizing that factoring out common terms is more efficient than expanding. Students who miss this strategic insight either work much harder than necessary or fall into the trap of thinking common factors automatically mean infinite solutions.