AMC 10 · 2017 · #23

Grade 8 algebra
polynomial-rootspolynomial-factoringvieta-formulas polynomial-factoringvieta-formulas ↑ Prerequisites: polynomial-roots
📏 Medium solution 💡 3 insights
Problem
Every root of a cubic is also a root of a quartic. Evaluate the quartic at one point.

Pick an answer.

(A)
-9009
(B)
-8008
(C)
-7007
(D)
-6006
(E)
-5005
How to solve
Strategy Introduce a Variable

f shares all three roots of g and has exactly one more root. Since both are monic, f must equal g times one linear factor (x - r). Name that fourth root r, expand the product, then match it term-by-term against the given f. Matching coefficients turns the problem into small linear equations that pin down r, a, b, and c.

1STEP 1

Write f as g times a linear factor

The quartic is the cubic times a linear factor.

f(x) = (x³ + ax² + x + 10)(x - r)
2STEP 2

Expand the product

Expanding writes every coefficient explicitly.

f(x) = x⁴ + (a - r)x³ + (1 - ar)x² + (10 - r)x - 10r
3STEP 3

Match coefficients to find r

One coefficient names the extra root.

10 - r = 100 → r = -90
4STEP 4

Find a, c, then b

The rest then follow immediately.

a = -89, c = 900, b = 1 - (-89)(-90) = -8009
5STEP 5

Evaluate f(1)

Evaluating gives -7007, choice (B).

f(1) = 2 - 8009 + 100 + 900 = -7007
Answer
-7007
Two independent routes agree: summing coefficients gives 2 - 8009 + 100 + 900 = -7007, and the factored form gives g(1)(1 - r) = (-77)(91) = -7007. The factored route never needs b at all, so it cross-checks the coefficient matching. -7007 is exactly choice (C).
💡Key takeaway

When one monic polynomial holds all the roots of a smaller monic one, it is just the smaller one times a single (x - r) leftover, and matching coefficients reveals everything.

  • Write f as g times a linear factor
  • Expand the product
  • Match coefficients to find r
  • Find a, c, then b
  • Evaluate f(1)