2003 AIME II Problem 9

Attempt Problem 9 of the 2003 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AIME II solutions, or check the answer key.

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9.

Consider the polynomials

P(x) = x^6 - x^5 - x^3 - x^2 - x

and

Q(x) = x^4 - x^3 - x^2 - 1.

Given that

z_1,

z_2,

z_3,

and

z_4

are the roots of

Q(x) = 0,

find

P(z_1) + P(z_2) + P(z_3) + P(z_4).

Answer

Answer: 6

Concepts:polynomial Vieta’s Formulas

Difficulty rating: 2400

Small hint:

Divide

P(x)

by

Q(x);

at the roots of

Q,

only the remainder survives

Big hint:

P(x) = Q(x)(x^2 + 1)

+ x^2 - x + 1,

so sum

z_i^2 - z_i + 1

using Vieta’s formulas on

Q

Solution:

Polynomial division gives

\begin{aligned} &P(x) = Q(x)\,(x^2 + 1) \\ &\quad {}+ x^2 - x + 1, \end{aligned}

so

P(z_i) = z_i^2 - z_i + 1

for each root

z_i

of

Q.

By Vieta’s formulas for

Q(x) = x^4 - x^3 - x^2 - 1,

we have

\sum z_i = 1

and

\sum_{i \lt j} z_i z_j = -1,

so

\sum z_i^2 = \left(\sum z_i\right)^2 - 2\sum_{i \lt j} z_i z_j

= 1 + 2 = 3.

Therefore

\sum_{i=1}^4 P(z_i) = 3 - 1 + 4 = 6.