Question:The function f is defined for all x neq -2 by the equation \(\mathrm{f(x) = \frac{(x^2 - 16)(x + 5)}{(x...

1. TRANSLATE the problem information

• Given: \(\mathrm{f(x) = \frac{(x^2 - 16)(x + 5)}{(x + 2)}}\) for all \(\mathrm{x \neq -2}\)
• Given: \(\mathrm{f(2k - 3) = 0}\) where k is a constant
• Find: Product of all possible values of k

2. INFER when a rational function equals zero

• A rational function equals zero only when its numerator equals zero AND its denominator does not equal zero
• Since we need \(\mathrm{f(2k - 3) = 0}\), the input value \(\mathrm{(2k - 3)}\) must make the numerator zero while keeping the denominator non-zero

3. Find the zeros of the function

• Set the numerator equal to zero: \(\mathrm{(x^2 - 16)(x + 5) = 0}\)
• SIMPLIFY using the zero product property:
  • \(\mathrm{x^2 - 16 = 0}\) OR \(\mathrm{x + 5 = 0}\)
  • \(\mathrm{x^2 = 16}\) OR \(\mathrm{x = -5}\)
  • \(\mathrm{x = \pm 4}\) OR \(\mathrm{x = -5}\)
• The zeros are \(\mathrm{x = 4}\), \(\mathrm{x = -4}\), and \(\mathrm{x = -5}\)

4. APPLY CONSTRAINTS to verify valid zeros

• Check that none of these zeros make the denominator \(\mathrm{(x + 2)}\) equal to zero
• Since \(\mathrm{x + 2 = 0}\) when \(\mathrm{x = -2}\), and none of our zeros equal \(\mathrm{-2}\), all zeros are valid

5. CONSIDER ALL CASES for the equation \(\mathrm{f(2k - 3) = 0}\)

• Since \(\mathrm{f(2k - 3) = 0}\), we need \(\mathrm{2k - 3}\) to equal one of the zeros:
  • Case 1: \(\mathrm{2k - 3 = 4}\)
    \(\mathrm{2k = 7}\)
    \(\mathrm{k = \frac{7}{2}}\)
  • Case 2: \(\mathrm{2k - 3 = -4}\)
    \(\mathrm{2k = -1}\)
    \(\mathrm{k = -\frac{1}{2}}\)
  • Case 3: \(\mathrm{2k - 3 = -5}\)
    \(\mathrm{2k = -2}\)
    \(\mathrm{k = -1}\)

6. APPLY CONSTRAINTS one more time

• Verify that none of our k values make \(\mathrm{2k - 3 = -2}\) (which would make the denominator zero)
• For \(\mathrm{2k - 3 = -2}\), we get \(\mathrm{k = \frac{1}{2}}\), which is not among our solutions
• All k values are valid

7. Calculate the final answer

• Product of all possible k values: \(\mathrm{\frac{7}{2} \times (-\frac{1}{2}) \times (-1) = \frac{7}{4}}\) (use calculator)

Answer: \(\mathrm{\frac{7}{4}}\) or \(\mathrm{1.75}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the connection between \(\mathrm{f(2k-3) = 0}\) and finding where the original function equals zero. Instead, they try to substitute \(\mathrm{2k-3}\) directly into the function and set the entire expression equal to zero, leading to a much more complicated equation. This causes confusion and often leads to guessing rather than systematic solution.

Second Most Common Error:

Incomplete CONSIDER ALL CASES reasoning: Students find one or two zeros of the function but miss the third case. For example, they might solve \(\mathrm{x^2 - 16 = 0}\) to get \(\mathrm{x = \pm 4}\), but forget that \(\mathrm{x + 5 = 0}\) gives \(\mathrm{x = -5}\). This leads them to calculate the product of only two k values instead of three, giving an incorrect final answer.

The Bottom Line:

This problem requires recognizing that the composite function \(\mathrm{f(2k-3)}\) equals zero exactly when its input equals a zero of the original function. Students who miss this key insight often get bogged down in unnecessary algebra and lose sight of the systematic approach needed to find all possible solutions.