Question:Which of the following expressions are factors of 2x^3 - 5x^2 - 14x + 8?\((\mathrm{x} - 4)\)\((\mathrm{x} + 2)\)

1. TRANSLATE the factor expressions to testable values

• Given information:
  • Polynomial: \(2\mathrm{x}^3 - 5\mathrm{x}^2 - 14\mathrm{x} + 8\)
  • Test expressions: \((\mathrm{x} - 4)\) and \((\mathrm{x} + 2)\)

• What this tells us: We need to check if these are factors using the Factor Theorem

2. INFER the testing strategy

• Key insight: By the Factor Theorem, \((\mathrm{x} - \mathrm{a})\) is a factor if and only if \(\mathrm{P(a)} = 0\)
• For \((\mathrm{x} - 4)\): test \(\mathrm{x} = 4\)
• For \((\mathrm{x} + 2)\): this equals \((\mathrm{x} - (-2))\), so test \(\mathrm{x} = -2\)

3. SIMPLIFY by evaluating P(4)

Test if \((\mathrm{x} - 4)\) is a factor:

\(\mathrm{P(4)} = 2(4)^3 - 5(4)^2 - 14(4) + 8\)
\(= 2(64) - 5(16) - 56 + 8\)
\(= 128 - 80 - 56 + 8\)
\(= 0\)

Since \(\mathrm{P(4)} = 0\), \((\mathrm{x} - 4)\) is a factor ✓

4. SIMPLIFY by evaluating P(-2)

Test if \((\mathrm{x} + 2)\) is a factor:

\(\mathrm{P(-2)} = 2(-2)^3 - 5(-2)^2 - 14(-2) + 8\)
\(= 2(-8) - 5(4) + 28 + 8\)
\(= -16 - 20 + 28 + 8\)
\(= 0\)

Since \(\mathrm{P(-2)} = 0\), \((\mathrm{x} + 2)\) is also a factor ✓

5. INFER the final answer

Both expressions are factors, so the answer is "I and II"

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students often mishandle \((\mathrm{x} + 2)\) by substituting \(\mathrm{x} = 2\) instead of \(\mathrm{x} = -2\). They don't recognize that \((\mathrm{x} + 2) = (\mathrm{x} - (-2))\), so they test the wrong value.

When they substitute \(\mathrm{P(2)}\):

\(\mathrm{P(2)} = 2(8) - 5(4) - 14(2) + 8\)
\(= 16 - 20 - 28 + 8\)
\(= -24 ≠ 0\)

Since this doesn't equal zero, they conclude \((\mathrm{x} + 2)\) is not a factor, leading them to select Choice A (I only).

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when working with negative numbers, particularly with \((-2)^3\) vs \((-2)^2\). A common mistake is computing \((-2)^3 = +8\) instead of \(-8\), or mixing up the signs in the final calculation.

This leads to getting \(\mathrm{P(-2)} ≠ 0\) when it should equal 0, causing them to incorrectly conclude that \((\mathrm{x} + 2)\) is not a factor and select Choice A (I only).

The Bottom Line:

This problem tests both conceptual understanding of the Factor Theorem and careful execution with negative number arithmetic. The key insight is recognizing that \((\mathrm{x} + 2)\) requires testing \(\mathrm{x} = -2\), not \(\mathrm{x} = 2\).