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)?

22. Since R(x) has degree three, then P(x) . 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. 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. Clearly, we could not have included any other constant functions. For any linear function, we have 2.Q(2) = Q(1) + Q(3) because 2(2)=1+3. So we have not included any other linear functions. Therefore, the desired answer is 27 - 5 = 22.