Polynomials

Algebra Level 5

Let P ( x ) = ( x 3 ) ( x 4 ) ( x 5 ) P(x) = (x - 3)(x - 4)(x - 5) . For how many polynomials Q ( x ) Q(x) , does there exist a polynomial R ( x ) R(x) of degree 3 such that P ( Q ( x ) ) = P ( x ) R ( x ) P(Q(x)) = P(x)R(x) ?


The answer is 22.

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1 solution

Mark Hennings
Aug 15, 2017

We want Q ( x ) Q(x) to be a quadratic polynomial (so that P ( Q ( x ) ) P(Q(x)) has degree 6 6 ). For P ( x ) P(x) to divide P ( Q ( x ) ) P(Q(x)) , we want P ( Q ( x ) ) P(Q(x)) to have zeros at 3 , 4 , 5 3,4,5 , and hence each of Q ( 3 ) , Q ( 4 ) , Q ( 5 ) Q(3),Q(4),Q(5) must belong to { 3 , 4 , 5 } \{3,4,5\} .

There is a unique polynomial Q a , b , c ( x ) Q_{a,b,c}(x) of degree at most 2 2 such that Q a , b , c ( 3 ) = a Q_{a,b,c}(3)=a , Q a , b , c ( 4 ) = b Q_{a,b,c}(4)=b and Q a , b , c ( 5 ) = c Q_{a,b,c}(5) = c , and so each valid Q ( x ) Q(x) must be a quadratic Q a , b , c ( x ) Q_{a,b,c}(x) where a , b , c { 3 , 4 , 5 } a,b,c \in \{3,4,5\} . Considering the 27 27 polynomials Q a , b , c ( x ) Q_{a,b,c}(x) for a , b , c { 3 , 4 , 5 } a,b,c \in \{3,4,5\} , the polynomial Q a , b , c ( x ) Q_{a,b,c}(x) will have degree 2 2 except when a , b , c a,b,c are in arithmetic progression (in which case Q a , b , c ( x ) Q_{a,b,c}(x) will be linear or constant). Only the triples ( 3 , 3 , 3 ) , ( 4 , 4 , 4 ) , ( 5 , 5 , 5 ) , ( 3 , 4 , 5 ) , ( 5 , 4 , 3 ) (3,3,3)\,,\,(4,4,4)\,,\,(5,5,5)\,,\,(3,4,5)\,,\,(5,4,3) are in arithmetic progression, and so there are 22 \boxed{22} polynomials Q ( x ) Q(x) .

Wow, I didn't realize this problem was so easy. Your solution took like no work. It was so simple to understand too. You pretty much only had to use the fact that a parabola can't share three points with a straight line. I came up with a method that worked, but it essentially involved solving for all the possible polynomials. I was able to get 5 of them without considering imaginary roots. I kinda gave up on it though. Here's a description of the method for anyone who's interested. The first step was always to determine the middle term of the quadratic factors based on an assumed layout of the three known factors.This helped me first determine at least one quadratic factor had to contain 2 of the factors (x-3), (x-4), (x-5)) because an assumption otherwise led to a set of inconsistent linear equations. Then since the third factor had to show up somewhere else, and the middle term had to be the same in all three quadratic factors, the spacing of the roots of the quadratic factors could be reasoned out fairly easily because two quadratic factors could immediately be found (up to a constant multiple) in most cases. Then using the fact that the constant term had to differ by one or two allowed me to find the leading coefficient. The constant term was also easily calculated from the leading term. Each combination of two factors produced a different set of polynomials. This wasn't the easiest method, but it was the easiest brute force method I could come up with.

James Wilson - 3 years, 9 months ago

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