Function h is defined by \(\mathrm{h(x) = (x - 4)(x + 1)(x + 5)}\). Function g is defined by \(\mathrm{g(x)...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{h(x) = (x - 4)(x + 1)(x + 5)}\)
  • \(\mathrm{g(x) = h(2x - 3)}\)
  • Need to find sum of x-intercepts of \(\mathrm{g(x)}\)

2. INFER the connection between zeros

• Key insight: X-intercepts occur where \(\mathrm{g(x) = 0}\)
• Since \(\mathrm{g(x) = h(2x - 3)}\), we need \(\mathrm{h(2x - 3) = 0}\)
• This happens when \(\mathrm{2x - 3}\) equals a zero of \(\mathrm{h(x)}\)

3. INFER the zeros of h(x)

• From \(\mathrm{h(x) = (x - 4)(x + 1)(x + 5)}\), the zeros are:
  • \(\mathrm{x = 4}\) (from \(\mathrm{x - 4 = 0}\))
  • \(\mathrm{x = -1}\) (from \(\mathrm{x + 1 = 0}\))
  • \(\mathrm{x = -5}\) (from \(\mathrm{x + 5 = 0}\))

4. SIMPLIFY to find x-intercepts of g(x)

• Set \(\mathrm{2x - 3}\) equal to each zero of \(\mathrm{h(x)}\):

For \(\mathrm{2x - 3 = 4}\):
\(\mathrm{2x = 7}\)
\(\mathrm{x = \frac{7}{2}}\)

For \(\mathrm{2x - 3 = -1}\):
\(\mathrm{2x = 2}\)
\(\mathrm{x = 1}\)

For \(\mathrm{2x - 3 = -5}\):
\(\mathrm{2x = -2}\)
\(\mathrm{x = -1}\)

5. SIMPLIFY the final sum

• \(\mathrm{a + b + c = \frac{7}{2} + 1 + (-1)}\)
  \(\mathrm{= \frac{7}{2} + 0}\)
  \(\mathrm{= \frac{7}{2}}\)

Answer: C) 7/2

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students attempt to expand \(\mathrm{g(x) = h(2x - 3)}\) by substituting and multiplying out all the terms, creating a complex cubic expression instead of using the zero-finding approach.

They might try to expand \(\mathrm{[(2x - 3) - 4][(2x - 3) + 1][(2x - 3) + 5] = (2x - 7)(2x - 2)(2x + 2)}\), then attempt to find zeros of this expanded form. This leads to much more complicated algebra and increases the chance of calculation errors, potentially causing them to abandon the systematic solution and guess.

Second Most Common Error:

Poor SIMPLIFY execution: Students understand the correct approach but make arithmetic errors when solving the linear equations or when adding the final sum.

For example, they might incorrectly solve \(\mathrm{2x - 3 = 4}\) as \(\mathrm{x = \frac{1}{2}}\) instead of \(\mathrm{x = \frac{7}{2}}\), or make sign errors when calculating the final sum. This may lead them to select Choice A (-2) or Choice B (-1).

The Bottom Line:

The key insight is recognizing that function composition problems involving zeros can often be solved by working backwards from the inner function's zeros, rather than expanding everything algebraically.