2005 AMC 12A Problems/Problem 24

Problem

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 $P(Q(x))=P(x)* R(x)$ ?

Solution

Since $R(x)$ has degree three, then $P(x)\cdot R(x)$ has degree six. Thus, $P(Q(x))$ has degree six, so $Q(x)$ must have degree two, since $P(x)$ has degree three.

$P(Q(1))=(Q(1)-1)(Q(1)-2)(Q(1)-3)=P(1)\cdot R(1)=0,$
$P(Q(2))=(Q(2)-1)(Q(2)-2)(Q(2)-3)=P(2)\cdot R(2)=0,$
$P(Q(3))=(Q(3)-1)(Q(3)-2)(Q(3)-3)=P(3)\cdot R(3)=0.$

Hence, we conclude $Q(1)$ , $Q(2)$ , and $Q(3)$ must each be $1$ , $2$ , or $3$ . Since a quadratic is uniquely determined by three points, there can be $3*3*3 = 27$ different quadratics $Q(x)$ after each of the values of $Q(1)$ , $Q(2)$ , and $Q(3)$ are chosen.

However, we have included $Q(x)$ which are not quadratics. Namely,

$Q(1)=Q(2)=Q(3)=1 \Rightarrow Q(x)=1,$
$Q(1)=Q(2)=Q(3)=2 \Rightarrow Q(x)=2,$
$Q(1)=Q(2)=Q(3)=3 \Rightarrow Q(x)=3,$
$Q(1)=1, Q(2)=2, Q(3)=3 \Rightarrow Q(x)=x,$
$Q(1)=3, Q(2)=2, Q(3)=1 \Rightarrow Q(x)=4-x.$

Clearly, we could not have included any other constant functions. For any linear function, we have $2\cdot Q(2) = Q(1) + Q(3)$ . Again, it is pretty obvious that we have not included any other linear functions. Therefore, the desired answer is $27 - 5 = 22$ .

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2005 AMC 12A Problems/Problem 24

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