Help me!

Let P(x)=(x1)(x2)(x3)P(x) = (x-1)(x-2)(x-3).

Find 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) \cdot R(x).

Note by Vedant Sharda
3 years, 9 months ago

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What are your thoughts? What have you tried?

Must the polynomial have complex, real or integer coefficients?

Calvin Lin Staff - 3 years, 9 months ago

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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 ?

Vedant Sharda - 3 years, 9 months ago

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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(1) ? Q(2)? Q(3)?

Calvin Lin Staff - 3 years, 9 months ago

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@Calvin Lin To find Q(1) , put x=1 in

(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

Vedant Sharda - 3 years, 9 months ago

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@Vedant Sharda Great, so we now know the possible values of Q(1),Q(2),Q(3) Q(1), Q(2), Q(3) .

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

Calvin Lin Staff - 3 years, 9 months ago

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@Calvin Lin I can't get it

Vedant Sharda - 3 years, 9 months ago

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@Vedant Sharda Check out lagrange interpolation

Hint: If we have nn equations of the form f(xi)=yi f(x_i) = y _ i , then there is a unique polynomial of degree at most n1n-1 which satisfies the conditions.

Calvin Lin Staff - 3 years, 9 months ago

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I think the answer is 2

Aman Pathak - 3 years, 9 months ago

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Answer is 22

Vedant Sharda - 3 years, 9 months ago

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

Calvin Lin Staff - 3 years, 9 months ago

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Let Q(x)=ax2+bx+cQ(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.

Alan Yan - 3 years, 8 months ago

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