AMC 10 · 2024 · #15

Grade 11 algebra
polynomial-rootscomplex-numbersdifference-of-squarespolynomial-factoringvieta-formulas convert-to-algebraeasier-related-problem ↑ Prerequisites: polynomial-factoringdifference-of-squarescomplex-numbers
📏 Long solution 💡 4 insights
Problem
The cubic x³ + 2x² - x + 3 has three roots, named p, q, and r. Find the number (p² + 4)(q² + 4)(r² + 4).

Pick an answer.

(A)
64
(B)
75
(C)
100
(D)
125
(E)
144
How to solve
Strategy Organize Information in More Ways

The obvious move is to solve the cubic and plug the roots in. Tool #16 (Change Focus) kills that idea early: a quick rational-root test shows the roots are not nice, so the answer cannot come from the roots individually — it has to come from the polynomial as a whole. Tool #13 (Convert to Algebra) supplies the bridge, turning the sentence "p, q, r are the roots" into the identity f(x) = (x-p)(x-q)(x-r), which reports the product of any input's distances to all three roots at once. Then Tool #15 (Organize Information in More Ways) does the real work in two passes: first it rewrites each p²+4 as (p-2i)(p+2i), which is only possible once i is allowed, and then it re-sorts the resulting six factors into two piles that each match the factored polynomial. The catch is that each pile sits backwards relative to f, so three minus signs come out of each pile — the sign bookkeeping is the entire difficulty here. Tool #9 (Solve an Easier Related Problem) audits that sign on a cubic whose roots are visible, before the real numbers get substituted.

1STEP 1

Check whether the roots are findable

Every rational candidate fails, so do not chase roots.

f(1) = 5, f(-1) = 5, f(3) = 45, f(-3) = -3
2STEP 2

Turn roots into a factored polynomial

The roots are just the factors of f.

f(x) = x³ + 2x² - x + 3 = (x-p)(x-q)(x-r)
3STEP 3

Split using i

Square-plus-four becomes two factors using 2i.

p² + 4 = p² - (2i)² = (p - 2i)(p + 2i)
4STEP 4

Regroup six factors into two

The product folds into f at 2i times f at -2i.

(p²+4)(q²+4)(r²+4) = (-f(2i))(-f(-2i)) = f(2i) f(-2i)
5STEP 5

Audit the sign on a solvable cubic

A toy cubic confirms the sign is right.

4 · 5 · 5 = 100 = (-10i)(10i) = g(2i) g(-2i)
6STEP 6

Evaluate at 2i and -2i

The two values are conjugates.

f(2i) = -5 - 10i, f(-2i) = -5 + 10i
7STEP 7

Multiply the conjugate pair

Their product is the real number 125.

f(2i) f(-2i) = (-5 - 10i)(-5 + 10i) = 25 - 100i² = 125
Answer
125
Check numerically against the roots the solution deliberately avoided. The cubic has one real root near -2.7573 and a conjugate pair near 0.3786 ± 0.9719i. For the real root, p² + 4 ≈ 11.6026. For the pair, q² + 4 ≈ 3.1987 + 0.7360i and r² + 4 is its conjugate, so their product is 3.1987² + 0.7360² ≈ 10.7735. Multiplying, 11.6026 × 10.7735 ≈ 125.00 — the closed-form route landed exactly where brute force does. Two structural checks back this up. First, the result had to be a positive real: the non-real roots come in a conjugate pair, so their contributions multiply to a positive real, and the real root contributes p² + 4 > 0. Second, the choices flag the likely slips. Dropping one of the two minus signs in the regrouping step gives -125, which no choice offers. Misreading the target as p² - 4 and using f(2)f(-2) = 17 · 5 = 85 also misses the list. And 100 sitting in the list as (C) is the value of the toy cubic x³ - x used for the audit, so it is a trap for anyone who checks the method and then forgets to run the real polynomial.
💡Key takeaway

When a question is built out of a polynomial's roots but the roots are ugly, stop hunting for them: rewrite the request as the polynomial evaluated at a couple of well-chosen inputs, and count your minus signs on the way.

  • Check whether the roots are findable
  • Turn "roots" into a factored polynomial
  • Split p² + 4 using i
  • Regroup six factors and pay the signs
  • Audit the sign on a cubic you can solve
  • Evaluate f at 2i and -2i
  • Multiply the conjugate pair