Let \(\mathrm{f(x) = (x + 4)(x - 1)}\) and \(\mathrm{g(x) = (x - 4)(x + 1)}\). Suppose \(\mathrm{f(x) = g(x)...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{f(x) = (x + 4)(x - 1)}\) and \(\mathrm{g(x) = (x - 4)(x + 1)}\)
  • We need to solve \(\mathrm{f(x) = g(x) + 20}\)
• What this tells us: We need to expand both functions and set up an equation

2. SIMPLIFY by expanding both quadratic expressions

• Expand f(x):
  \(\mathrm{f(x) = (x + 4)(x - 1)}\)
  \(\mathrm{= x^2 - x + 4x - 4}\)
  \(\mathrm{= x^2 + 3x - 4}\)

• Expand g(x):
  \(\mathrm{g(x) = (x - 4)(x + 1)}\)
  \(\mathrm{= x^2 + x - 4x - 4}\)
  \(\mathrm{= x^2 - 3x - 4}\)

3. TRANSLATE the condition into an equation

• Set up: \(\mathrm{f(x) = g(x) + 20}\)
• Substitute: \(\mathrm{x^2 + 3x - 4 = (x^2 - 3x - 4) + 20}\)
• SIMPLIFY: \(\mathrm{x^2 + 3x - 4 = x^2 - 3x + 16}\)

4. SIMPLIFY the equation step by step

• Subtract \(\mathrm{x^2}\) from both sides: \(\mathrm{3x - 4 = -3x + 16}\)
• Add 3x to both sides: \(\mathrm{6x - 4 = 16}\)
• Add 4 to both sides: \(\mathrm{6x = 20}\)
• Divide by 6: \(\mathrm{x = \frac{20}{6} = \frac{10}{3}}\)

Answer: D (10/3)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Sign errors during binomial expansion, particularly with \(\mathrm{(x - 4)(x + 1)}\)

Students often expand \(\mathrm{g(x) = (x - 4)(x + 1)}\) as \(\mathrm{x^2 + 4x - x - 4 = x^2 + 3x - 4}\) instead of the correct \(\mathrm{x^2 - 3x - 4}\). This happens because they lose track of the negative sign in front of the 4. With this error, they would get \(\mathrm{f(x) = g(x)}\), leading to \(\mathrm{0 = 20}\), which creates confusion about whether there's no solution or they made an error somewhere.

This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY reasoning: Arithmetic mistakes when solving the linear equation \(\mathrm{6x = 20}\)

Some students correctly set up the equation but make calculation errors, perhaps getting \(\mathrm{x = \frac{20}{3}}\) instead of \(\mathrm{x = \frac{10}{3}}\), or incorrectly reducing the fraction. Since \(\mathrm{\frac{20}{3} \approx 6.67}\) doesn't match any answer choice exactly, this confusion might lead them to select Choice C (1) as the "closest looking" integer answer.

The Bottom Line:

This problem tests your ability to carefully track signs through multiple algebraic steps. The key insight is that while the algebra looks intimidating with two quadratic expressions, the \(\mathrm{x^2}\) terms cancel out completely, leaving you with a simple linear equation to solve.