AMC 10 · 2022 · #15
Grade 8 geometry-3dPick an answer.
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.
Expand the new volume
Expanding leaves only symmetric sums.
Every product of three binomials in a, b, c splits into pieces that only know the symmetric sums — perfect for Vieta.
6.EE.A.3Identify SubproblemsRead them off the coefficients
Read them off the coefficients without solving.
Vieta turns coefficients into root-sums — no root-finding required.
The coefficients hand over the sums and products of the roots directly, with no root-finding needed.
▸ Why?
A polynomial's coefficients record exactly those symmetric combinations of its roots.
▸ Why?
Expanding the new product splits it into pieces that only ever mention those same combinations.
Substitute and compute
Substituting gives the volume.
Common denominator 10 lets us add four fractions in one shot.
5.NF.A.1Identify SubproblemsMatch the choice
The volume is 30.
Match the computed value to the listed options.
4.NBT.A.2Eliminate PossibilitiesWe 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 6/10 + 58/10 + 156/10 + 80/10 = 30 — choice (D).