Let r, s, and t be the solutions to the equation 2x^3 - 5x^2 - 14x + 40 = 0....

1. TRANSLATE the problem information

• Given information:
  • Cubic equation: \(\mathrm{2x^3 - 5x^2 - 14x + 40 = 0}\)
  • r, s, and t are the three solutions (roots)
  • Need to find: rst (product of all three roots)

2. INFER the most efficient approach

• Key insight: Rather than solving for individual roots and multiplying them, we can use Vieta's formulas
• Vieta's formulas give us direct relationships between coefficients and products/sums of roots
• For a cubic equation \(\mathrm{ax^3 + bx^2 + cx + d = 0}\) with roots r, s, t:
  • \(\mathrm{rst = -d/a}\) (this is exactly what we need!)

3. TRANSLATE the equation to identify coefficients

• From \(\mathrm{2x^3 - 5x^2 - 14x + 40 = 0}\):
  • \(\mathrm{a = 2}\) (coefficient of x³)
  • \(\mathrm{b = -5}\) (coefficient of x²)
  • \(\mathrm{c = -14}\) (coefficient of x)
  • \(\mathrm{d = 40}\) (constant term)

4. SIMPLIFY using Vieta's formula

• Apply \(\mathrm{rst = -d/a}\):
  • \(\mathrm{rst = -40/2 = -20}\)

Answer: (B) -20

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that Vieta's formulas apply here and instead attempt to solve the cubic equation completely.

They might try factoring or using the cubic formula to find individual values of r, s, and t, then multiply them together. This approach is extremely time-consuming and error-prone, especially since the cubic doesn't factor easily. Students often abandon this approach partway through or make calculation errors.

This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Conceptual confusion about Vieta's formulas: Students remember that Vieta's formulas exist but apply the wrong relationship.

They might use \(\mathrm{rst = d/a}\) (forgetting the negative sign) and calculate \(\mathrm{40/2 = 20}\), leading them to select Choice (D) (20).

The Bottom Line:

This problem tests whether students recognize when a direct formula (Vieta's) is more efficient than brute-force calculation. The key insight is that finding products and sums of roots is often easier than finding the individual roots themselves.