2005 AMC 12A Problem 24

Question

Let

P(x) = (x - 1)(x - 2)(x - 3).

For how many polynomials

Q(x)

does there exist a polynomial

R(x)

of degree

3

such that

P(Q(x)) = P(x) \cdot R(x)?

19

22

24

27

32

Answer: B

Concepts:polynomial casework

Difficulty rating: 2520

Small hint:

Compare degrees:

P(Q(x))

has degree

3\deg Q

and

P(x)R(x)

has degree

6,

so

\deg Q = 2

Big hint:

For

x = 1, 2, 3,

the right side is

0,

so each of

Q(1), Q(2), Q(3)

lies in

\{1, 2, 3\};

discard the triples giving degree less than

2

Solution:

Since

P(x)R(x)

has degree

6

and

P(Q(x))

has degree

3\deg Q,

we need

\deg Q = 2.

A quadratic

Q

is determined by the ordered triple

(Q(1), Q(2), Q(3)).

At

x = 1, 2, 3

the right side vanishes, so

P(Q(x)) = 0,

forcing each of

Q(1), Q(2), Q(3)

into

\{1, 2, 3\}.

That gives

27

triples.

Five of them give a polynomial of degree less than

2:

the constants from

(1,1,1), (2,2,2), (3,3,3)

and the linear

Q(x) = x

from

(1,2,3)

and

Q(x) = 4 - x

from

(3,2,1).

The other

27 - 5 = 22

triples are non-collinear and yield genuine quadratics.

Thus, the correct answer is B.

2005 AMC 12A Problem 24

Answer: B

Difficulty rating: 2520

Small hint:

Big hint:

Solution:

Problem 24 in other years