Which expression is equivalent to \((\mathrm{d} - 6)(8\mathrm{d}^2 - 3)\)?
GMAT Advanced Math : (Adv_Math) Questions
Which expression is equivalent to \((\mathrm{d} - 6)(8\mathrm{d}^2 - 3)\)?
1. INFER the approach needed
- We have two expressions being multiplied: \(\mathrm{(d - 6)}\) and \(\mathrm{(8d^2 - 3)}\)
- This requires using the distributive property to multiply each term in the first expression by each term in the second
2. SIMPLIFY using the distributive property (first application)
- Distribute each term from \(\mathrm{(d - 6)}\):
- \(\mathrm{(d - 6)(8d^2 - 3) = d(8d^2 - 3) - 6(8d^2 - 3)}\)
3. SIMPLIFY using the distributive property (second application)
- Now distribute within each grouped term:
- \(\mathrm{d(8d^2 - 3) = d(8d^2) + d(-3) = 8d^3 - 3d}\)
- \(\mathrm{-6(8d^2 - 3) = -6(8d^2) + (-6)(-3) = -48d^2 + 18}\)
4. SIMPLIFY by combining all terms
- Put all terms together: \(\mathrm{8d^3 - 3d - 48d^2 + 18}\)
- Arrange in standard form: \(\mathrm{8d^3 - 48d^2 - 3d + 18}\)
Answer: C. \(\mathrm{8d^3 - 48d^2 - 3d + 18}\)
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY execution: Students make calculation errors when multiplying \(\mathrm{-6}\) by \(\mathrm{8d^2}\), getting \(\mathrm{-14d^2}\) instead of \(\mathrm{-48d^2}\).
This happens when students incorrectly calculate \(\mathrm{-6 \times 8 = -14}\) instead of \(\mathrm{-48}\), possibly confusing this with \(\mathrm{-6 - 8 = -14}\).
This may lead them to select Choice A (\(\mathrm{8d^3 - 14d^2 - 3d + 18}\))
Second Most Common Error:
Poor INFER reasoning: Students don't systematically apply the distributive property to all terms, missing some multiplications entirely.
They might only distribute the first term (d) or make incomplete distributions, leading to missing terms or wrong coefficients in their final answer.
This causes them to get stuck and guess among the remaining choices.
The Bottom Line:
This problem requires careful, systematic application of the distributive property with attention to sign management and arithmetic accuracy. Success depends on methodically multiplying each term by each term without skipping steps.