AMC 10 · 2022 · #16
Grade 8 geometry-3dProblem
The roots of the polynomial 10x3−39x2+29x−6 are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by 2
units. What is the volume of the new box?
Pick an answer.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The roots $a$, $b$, $c$ of the cubic $10x^3 - 39x^2 + 29x - 6$ are the three edge lengths of a rectangular box. Each edge is then lengthened by $2$. Find the volume of the new box.
Givens: Polynomial: $10x^3 - 39x^2 + 29x - 6$; Its roots $a, b, c$ are the original box dimensions; New edges: $a+2, b+2, c+2$; Choices: (A) $\tfrac{24}{5}$, (B) $\tfrac{42}{5}$, (C) $\tfrac{81}{5}$, (D) $30$, (E) $48$
Unknowns: New volume $V = (a+2)(b+2)(c+2)$
Understand
Restated: The roots $a$, $b$, $c$ of the cubic $10x^3 - 39x^2 + 29x - 6$ are the three edge lengths of a rectangular box. Each edge is then lengthened by $2$. Find the volume of the new box.
Givens: Polynomial: $10x^3 - 39x^2 + 29x - 6$; Its roots $a, b, c$ are the original box dimensions; New edges: $a+2, b+2, c+2$; Choices: (A) $\tfrac{24}{5}$, (B) $\tfrac{42}{5}$, (C) $\tfrac{81}{5}$, (D) $30$, (E) $48$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #13 Convert to Algebra, #3 Eliminate Possibilities
Two clean subproblems: (a) expand $(a+2)(b+2)(c+2)$ into symmetric sums of the roots, and (b) read those symmetric sums straight off the polynomial's coefficients via Vieta's relations. No actual root-finding needed — the trick is to see that the new volume depends only on $a+b+c$, $ab+bc+ca$, and $abc$, all of which Vieta hands us for free. Tool #7 names the split; Tool #13 lets us treat the roots as algebraic objects without computing them.
Execute — Answer: D
6.EE.A.3 Step 1 - Subproblem A: expand $(a+2)(b+2)(c+2)$ symbolically.
- Multiplying out and grouping like terms gives a sum that uses only the three elementary symmetric polynomials of $a, b, c$.
💡 Every product of three binomials in $a, b, c$ splits into pieces that only know the symmetric sums — perfect for Vieta.
8.EE.C.7 Step 2 - Subproblem B: read the symmetric sums off the polynomial $10x^3 - 39x^2 + 29x - 6$ using Vieta's relations.
- For $Ax^3 + Bx^2 + Cx + D$ with roots $a, b, c$: $a+b+c = -B/A$, $ab+bc+ca = C/A$, $abc = -D/A$.
💡 Vieta turns coefficients into root-sums — no root-finding required.
5.NF.A.1 Step 3 Substitute the three Vieta values into the expanded form from Step 1.
💡 Common denominator $10$ lets us add four fractions in one shot.
4.NBT.A.2 Step 4 - Match to the answer choices.
- $V = 30$ is choice (D).
💡 Match the computed value to the listed options.
6.EE.A.3 Subproblem A: expand $(a+2)(b+2)(c+2)$ symbolically. Multiplying out and groupin 8.EE.C.7 Subproblem B: read the symmetric sums off the polynomial $10x^3 - 39x^2 + 29x - 5.NF.A.1 Substitute the three Vieta values into the expanded form from Step 1. 4.NBT.A.2 Match to the answer choices. $V = 30$ is choice (D). Review
Reasonableness: Cross-check via the substitution shortcut: $P(x) = 10(x-a)(x-b)(x-c)$, so $P(-2) = 10(-2-a)(-2-b)(-2-c) = -10(a+2)(b+2)(c+2)$. Compute $P(-2) = 10(-8) - 39(4) + 29(-2) - 6 = -80 - 156 - 58 - 6 = -300$. Then $(a+2)(b+2)(c+2) = -P(-2)/10 = 300/10 = 30$ — matches choice (D). Magnitude is also plausible: with $abc = 0.6$ and the boosts of $+2$ on each edge, the dominant new term is $4(a+b+c) + 8 = 4 \cdot 3.9 + 8 = 23.6$, plus correction terms of about $6.4$ — total near $30$.
Alternative: Tool #5 (Look for a Pattern) combined with Tool #13 (Convert to Algebra): note that $P(-2) = 10(-2)^3 - 39(-2)^2 + 29(-2) - 6 = -80 - 156 - 58 - 6 = -300$. Since $P(x) = 10(x-a)(x-b)(x-c)$, we get $P(-2) = 10(-2-a)(-2-b)(-2-c) = -10(a+2)(b+2)(c+2)$. So $(a+2)(b+2)(c+2) = -P(-2)/10 = 300/10 = 30$. Same answer, one substitution.
CCSS standards used (min grade 8)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding the product $(a+2)(b+2)(c+2)$ symbolically into a sum of elementary symmetric polynomials in $a, b, c$.)8.EE.C.7Solve linear equations in one variable (Treating $a, b, c$ as algebraic roots of the polynomial and reading their symmetric sums off the coefficients via Vieta's formulas.)5.NF.A.1Add and subtract fractions with unlike denominators (Combining $\tfrac{6}{10} + \tfrac{58}{10} + \tfrac{156}{10} + \tfrac{80}{10}$ to get $\tfrac{300}{10} = 30$.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Matching the computed value $30$ to the answer choices.)
⭐ We never have to find the three roots themselves — Vieta hands us their sum, pair-sum, and product directly from the coefficients. Expanding $(a+2)(b+2)(c+2) = abc + 2(ab+bc+ca) + 4(a+b+c) + 8$ uses only those three numbers, and substituting gives $\tfrac{6}{10} + \tfrac{58}{10} + \tfrac{156}{10} + \tfrac{80}{10} = 30$ — choice $\textbf{(D)}$.
⭐ We never have to find the three roots themselves — Vieta hands us their sum, pair-sum, and product directly from the coefficients. Expanding $(a+2)(b+2)(c+2) = abc + 2(ab+bc+ca) + 4(a+b+c) + 8$ uses only those three numbers, and substituting gives $\tfrac{6}{10} + \tfrac{58}{10} + \tfrac{156}{10} + \tfrac{80}{10} = 30$ — choice $\textbf{(D)}$.