Question:The functions g and f are defined by \(\mathrm{g(x) = 2x^2 - 6x + 8}\) and \(\mathrm{f(x) = (x-3)g(x)}\). What...

1. TRANSLATE the problem requirement

• We need to find \(\mathrm{f(3)}\) where \(\mathrm{f(x) = (x-3)g(x)}\)
• This means substitute \(\mathrm{x = 3}\) into the function \(\mathrm{f(x)}\)

2. INFER the most efficient approach

• When we substitute \(\mathrm{x = 3}\), we get \(\mathrm{f(3) = (3-3)g(3)}\)
• Notice that \(\mathrm{(3-3) = 0}\)
• Since we're multiplying by 0, the result will be 0 no matter what \(\mathrm{g(3)}\) equals

3. SIMPLIFY the expression

• \(\mathrm{f(3) = (3-3)g(3)}\)
  \(\mathrm{= (0)g(3)}\)
  \(\mathrm{= 0}\)

Answer: B. 0

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skills: Students don't recognize the zero factor shortcut and instead calculate \(\mathrm{g(3)}\) first.

They compute \(\mathrm{g(3) = 2(3)^2 - 6(3) + 8}\)
\(\mathrm{= 18 - 18 + 8}\)
\(\mathrm{= 8}\), then multiply \(\mathrm{(0)(8) = 0}\). While this gives the correct answer, it's unnecessarily complicated and creates opportunities for arithmetic errors. If they make a calculation mistake finding \(\mathrm{g(3)}\), they might get confused about the final answer even though the zero factor makes \(\mathrm{g(3)}\) irrelevant.

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when calculating \(\mathrm{g(3)}\).

For example, they might calculate \(\mathrm{g(3) = 2(9) - 6(3) + 8}\)
\(\mathrm{= 18 - 18 + 8}\) incorrectly, getting \(\mathrm{g(3) = 11}\) instead of 8. Then \(\mathrm{f(3) = (0)(11) = 0}\) still gives the right answer, but if they're unsure about the zero multiplication, this could lead them to select Choice D (11).

The Bottom Line:

This problem tests whether students can spot algebraic shortcuts. The key insight is recognizing that a zero factor makes the entire product zero, eliminating the need for complex calculations.