The function f is defined by \(\mathrm{f(x) = x(x - 2)(x + 4)^2}\). The value of \(\mathrm{f(w(w - 3))}\) is...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{f(x) = x(x - 2)(x + 4)^2}\)
  • \(\mathrm{f(w(w - 3)) = 0}\)
  • Need to find sum of all possible w values

2. INFER the key relationship

• For \(\mathrm{f(w(w - 3)) = 0}\), the input \(\mathrm{w(w - 3)}\) must make f equal zero
• This means \(\mathrm{w(w - 3)}\) must equal one of the roots of \(\mathrm{f(x) = 0}\)
• Strategy: Find all roots of \(\mathrm{f(x)}\), then set \(\mathrm{w(w - 3)}\) equal to each root

3. SIMPLIFY to find the roots of f(x)

From \(\mathrm{f(x) = x(x - 2)(x + 4)^2 = 0}\):

• \(\mathrm{x = 0}\) (first factor)
• \(\mathrm{x = 2}\) (second factor)
• \(\mathrm{x = -4}\) (third factor, with multiplicity 2)

4. CONSIDER ALL CASES by setting w(w - 3) equal to each root

Case 1: \(\mathrm{w(w - 3) = 0}\)

• \(\mathrm{w^2 - 3w = 0}\)
• \(\mathrm{w(w - 3) = 0}\)
• Solutions: \(\mathrm{w = 0}\) or \(\mathrm{w = 3}\)

Case 2: \(\mathrm{w(w - 3) = 2}\)

• \(\mathrm{w^2 - 3w - 2 = 0}\)
• Using quadratic formula: \(\mathrm{w = \frac{3 ± \sqrt{9 + 8}}{2} = \frac{3 ± \sqrt{17}}{2}}\) (use calculator for \(\mathrm{\sqrt{17}}\))

Case 3: \(\mathrm{w(w - 3) = -4}\)

• \(\mathrm{w^2 - 3w + 4 = 0}\)
• Discriminant \(\mathrm{= (-3)^2 - 4(1)(4) = 9 - 16 = -7 \lt 0}\)
• No real solutions

5. SIMPLIFY the final sum

All possible w values: \(\mathrm{0, 3, \frac{3 + \sqrt{17}}{2}, \frac{3 - \sqrt{17}}{2}}\)

Sum \(\mathrm{= 0 + 3 + \frac{3 + \sqrt{17}}{2} + \frac{3 - \sqrt{17}}{2}}\)
\(\mathrm{= 0 + 3 + 3 + \frac{\sqrt{17}}{2} - \frac{\sqrt{17}}{2}}\)
\(\mathrm{= 6}\)

Answer: C) 6

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that \(\mathrm{f(w(w-3)) = 0}\) means \(\mathrm{w(w-3)}\) must equal a root of \(\mathrm{f(x) = 0}\). Instead, they try to substitute \(\mathrm{w(w-3)}\) directly into the expanded form of \(\mathrm{f(x)}\), leading to a complicated sixth-degree polynomial that's nearly impossible to solve by hand. This leads to confusion and guessing.

Second Most Common Error:

Poor CONSIDER ALL CASES execution: Students find the roots of \(\mathrm{f(x)}\) correctly but only consider one case (usually \(\mathrm{w(w-3) = 0}\) because it's easiest), missing the other valid w values from \(\mathrm{w(w-3) = 2}\). They calculate the sum as just \(\mathrm{0 + 3 = 3}\), leading them to select Choice B (3).

The Bottom Line:

This problem tests your ability to connect composite function zeros with the original function's roots. The key insight is recognizing that you don't need to expand everything - just use the fact that the inner function \(\mathrm{w(w-3)}\) must hit the outer function's zeros.