AMC 10 · 2022 · #16

Grade 8 geometry-3d
polynomial-rootsvieta-formulasvolume-rectangular-prism identify-subproblemsconvert-to-algebra ↑ Prerequisites: vieta-formulas
📏 Short solution 💡 2 insights
Problem
The roots a, b, c of the cubic 10x³ - 39x² + 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.

Pick an answer.

(A)
$\frac{24}{5}$
(B)
$\frac{42}{5}$
(C)
$\frac{81}{5}$
(D)
30
(E)
48

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

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.

1STEP 1

Expand (a+2)(b+2)(c+2) — every term regroups into abc + 2(ab+bc+ca) + 4(a+b+c) + 8, using only the roots' symmetric sums.

(a+2)(b+2)(c+2) = abc + 2(ab+bc+ca) + 4(a+b+c) + 8
2STEP 2

Vieta reads the sums straight off the coefficients: a+b+c = 3910\frac{39}{10}, ab+bc+ca = 2910\frac{29}{10}, abc = 610\frac{6}{10} — no root-finding.

a+b+c = 3910\frac{39}{10}, ab+bc+ca = 2910\frac{29}{10}, abc = 610\frac{6}{10}
3STEP 3

Substitute the Vieta values into Step 1's form: 6+58+156+8010\frac{6 + 58 + 156 + 80}{10} = 30010\frac{300}{10} = 30.

V = 610\frac{6}{10} + 2 · 2910\frac{29}{10} + 4 · 3910\frac{39}{10} + 8 = 6+58+156+8010\frac{6 + 58 + 156 + 80}{10} = 30010\frac{300}{10} = 30
4STEP 4

Match to the choices: V = 30 is choice (D).

V = 30 → (D)
Answer
30
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\frac{P(-2)}{10} = 30010\frac{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 · 3.9 + 8 = 23.6, plus correction terms of about 6.4 — total near 30.
💡Key takeaway

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 610\frac{6}{10} + 5810\frac{58}{10} + 15610\frac{156}{10} + 8010\frac{80}{10} = 30 — choice (D).