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# Help me!

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 $$P(Q(x)) = P(x) \cdot R(x)$$.

Note by Vedant Sharda
2 years, 1 month ago

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What are your thoughts? What have you tried?

Must the polynomial have complex, real or integer coefficients?

Staff - 2 years, 1 month ago

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 ?

- 2 years, 1 month ago

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)?

Staff - 2 years, 1 month ago

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

- 2 years, 1 month ago

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

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

Staff - 2 years, 1 month ago

I can't get it

- 2 years, 1 month ago

Check out lagrange interpolation

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

Staff - 2 years, 1 month ago

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.

- 2 years ago

I think the answer is 2

- 2 years, 1 month ago

- 2 years, 1 month ago

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

Staff - 2 years, 1 month ago