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)?
Answer Choices:
A. 19
B. 22
C. 24
D. 27
E. 32
Solution:
The polynomial P(x)â‹…R(x) has degree 6 , so Q(x) must have degree 2 . Therefore Q is uniquely determined by the ordered triple (Q(1),Q(2),Q(3)). When x=1,2, or 3 , we have 0=P(x)â‹…R(x)=P(Q(x)). It follows that (Q(1),Q(2),Q(3)) is one of the 27 ordered triples (i,j,k), where i,j, and k can\
be chosen from the set {1,2,3}. However, the choices (1,1,1),(2,2,2),(3,3,3), (1,2,3), and (3,2,1) lead to polynomials Q(x) defined by Q(x)=1,2,3,x, and 4−x, respectively, all of which have degree less than 2 . The other 22​ choices for ( Q(1),Q(2),Q(3)) yield non-collinear points, so in each case Q(x) is a quadratic polynomial.
The problems on this page are the property of the MAA's American Mathematics Competitions