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

Find for how many polynomials \(Q(x)\) does there exist a polynomial \(R(x)\) of degree 3 such that \(P(Q(x)) = P(x) \cdot R(x)\).

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## Comments

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TopNewestWhat are your thoughts? What have you tried?

Must the polynomial have complex, real or integer coefficients?

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What I have tried is :

Degree of P(x) is 3 and R(x) is also 3

So degree of P(x).R(x) is 6.

Degree of Q(x) must be 2 so that degree of P(Q(x)) is 6.

P(Q(x)) = P(x).R(x)

(Q(x)-1)(Q(x)-2)(Q(x)-3) = (x-1)(x-2)(x-3).R(x)

Now what I have to do ?

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Great job. A key piece of information is that the degree of Q(x) is 6.

What can we say about the value of \( Q(1) \)? Q(2)? Q(3)?

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(Q(x)-1)(Q(x)-2)(Q(x)-3) = (x-1)(x-2)(x-3).R(x)

[Q(1)-1][Q(1)-2][Q(1)-3] =0

Q(1)-1 = 0. So Q(1) =1

Q(1)-2 = 0. So Q(1) = 2

Q(1)-3 = 0. So Q(1) = 3

Hence Q(1) = 1or 2or 3

Similarly Q(2) = 1or 2or 3

And Q(3) = 1or 2or 3

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Lets say that \( Q(1) = 1, Q(2) = 2, Q(3) = 2 \). How many polynomials \( Q(x) \) of degree 2 satisfy that condition?

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lagrange interpolation

Check outHint:If we have \(n\) equations of the form \( f(x_i) = y _ i \), then there is a unique polynomial of degreeat most\(n-1\) which satisfies the conditions.Log in to reply

I think the answer is 2

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Answer is 22

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Yup, that's the numerical answer. The more interesting part is the actual solution.

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Let \(Q(x) = ax^2+bx + c \) and put that into P. Plug in x =1,2,3 and now you have a system of three linear equations with three possibilities each. Now it's just combinatorics! Calculate the inverse matrix and it is easy to see that out of the 27, 5 do not work. This the answer is 22.

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