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Remainder factor theorem: please help me reason this out

  1. Jan 23, 2017 #1
    1. The problem statement, all variables and given/known data
    find the number of polynomials f(x) that satisfies the condition:
    f(x) is monic polynomial, has degree 1000, has integer coefficients, and it can divide f(2x^3 + x)

    i would very much prefer that you guys give me hints first. thanks
    2. Relevant equations
    remainder factor theorem

    3. The attempt at a solution
    since f(2x^3 + x) = q(x)*f(x) such that q(x) is of degree 2000 and f(x) is of degree 1000 because f(2x^3 + x) is of degree 3000.

    since the problem asked for the number of f(x) that satisfies the condition, then i think it would be the number of ways to choose 1000 first degree polynomials in a group of 3000 first degree polynomials, which i think would be 3000C1000...? but wolfram alpha tells me the number has 829 digits which i think must be wrong because it doesn't fit the answer box that i'm trying to input my answer into.
     
  2. jcsd
  3. Jan 23, 2017 #2

    RUber

    User Avatar
    Homework Helper

    I don't have a complete solution at this point, but I think the fact that p is monic tells you to start with the leading coefficients. You should be able to constrain the possible q's based on that.
     
  4. Jan 23, 2017 #3
    i may have thought of that one too using rational root theorem, but the last term is not given so i don't know how to proceed from there.
     
  5. Jan 23, 2017 #4

    fresh_42

    Staff: Mentor

    My idea was to use ##f(g(x))'=f'(x)g'(x) = f'(x)q(x)+f(x)q'(x)## with ##g(x)=2x^3+x## and then to think about the leading and lowest terms.
     
  6. Jan 23, 2017 #5
    i'll give this one a go when i get up tomorrow. thanks
     
  7. Jan 23, 2017 #6

    fresh_42

    Staff: Mentor

    Not sure, whether it leads somewhere. I just thought it might be interesting.
     
  8. Jan 23, 2017 #7
     
  9. Jan 23, 2017 #8

    fresh_42

    Staff: Mentor

    Meanwhile, I think that leads nowhere. But the considerations of ##f(0), f(x)=x^n\cdot h(x)## with ##n\geq 0## and ##h(0)\neq 0## give the highest and lowest term of ##q(x)## and the consideration of all (complex) zeros yields a strong statement for these, i.e. the question whether ##f(a)=0## can hold for real ##a## beside ##a=0## and for which complex ##a##.
     
  10. Jan 24, 2017 #9
    i agree with your statement with f(x)= (x^n)*h(x) including f(0), following the idea.. what i was thinking is... can a thousand degree polynomial be factored into linear terms? since the question asked how many different polynomials, but never told us that it's factors cannot have non-integer coefficients.
     
  11. Jan 24, 2017 #10

    fresh_42

    Staff: Mentor

    I haven't considered the factorization, which of course could be done over ##\mathbb{C}##, even if we are not able to explicitly write down all ##1,000## roots as root expressions. I iterated ##f(g(x))=q(x)f(x)## instead. With ##f(x) =: x^nh(x)\, , \,h(0)\neq 0## we get ##(2x^2+1)^nh(g(x))=q(x)h(x)## which gives ##q(0)## and the highest term of ##q(x)##. Now ##f(z)=0## has a ##n-##fold zero at ##z=0## (##n\geq 0##) and the rest of its roots are the same as those of ##h(x)## which doesn't have a root at zero. But if ##h(z)=0##, what can you say about ##h(g(z))## and all other iterations with ##g(x)=x(2x^2+1)\,##? Can ##z## be real? What does it mean for complex solutions (roots) ##z=re^{i\varphi}## of ##h(x)##, resp. ##f(x)\,##? I haven't done it to the end, but it looks as these considerations give a lot of conditions, especially an investigation of ##r## looks promising.
     
  12. Jan 24, 2017 #11
    hmm... i thought we were saying the same thing, but i misunderstood f(x) = (x^n)(h(x)) give the highest and lowest degree of q(x). the way i understand it is... as n gets bigger then obviously the degree of q(x) must go lower to satisfy the degree of f(g(x)). However, you seem to be saying something different. i do not understand why ((2x^2 + 1)^n)*h(g(x)) = q(x)h(x). i also do not know what an n-fold means.
     
  13. Jan 24, 2017 #12
    this problem was lifted from an article on remainder factor theorem, but it seems much more complicated than that or am i just missing something here
     
  14. Jan 24, 2017 #13

    fresh_42

    Staff: Mentor

    Just substitute ##f(x)=x^nh(x)## in ##f(g(x))=f(x(2x^2+1))=q(x)f(x)##. Well, ##x=0## or in other words ##x\,\vert \,f(x)## or ##f(0)=0## is one possible solution. And we cannot say what the maximal ##n## with ##x^n\,\vert \,f(x)## is. So all ##n## between ##0## and ##1,000## are on principle possible. After we separated the maximal possible factor ##x^n## we can concentrate on ##h(x)## with ##h(0)\neq 0##.
     
  15. Jan 24, 2017 #14
    if f(x) = (x^n)(h(x)) and g(x) = x(2x+1) then f(g(x)) = (x^n)((2x^2 + 1)^n) * h(x(2x^2 + 1))?
     
  16. Jan 24, 2017 #15
    then i think we're saying almost the same thing. just in case we have n already, how do you think can we deduce the number of polynomials that are like f(x) from there?
     
  17. Jan 24, 2017 #16

    fresh_42

    Staff: Mentor

    Yes. I think it's easier to keep ##g(x)=x(2x^2+1)## where possible, so ##q(x)f(x)=q(x)x^nh(x)=f(g(x))=g(x)^nh(g(x))=x^n(2x^2+1)^nh(g(x))##.
    Now ##x^n## can be divided and we are left with ##h(x)## and ##h(g(x))\cdot (2x^2+1)^n=q(x)\cdot h(x)##.
    Here we can look for the highest and lowest terms of ##q(x)##. The lowest is easy, as it is ##h(g(0)) = h(0) = q(0) \cdot h(0)## and ##h(0) \neq 0##. For the highest some considerations on the degrees of the polynomials involved might be necessary. And we can investigate the solutions of ##h(z)=0##. First with ##z \in \mathbb{R}## and then with ##z=r\cdot e^{i\varphi}## for complex ##z##.

    I'm not sure, but it seems to be an interesting problem. Maybe there is a trick which I don't see yet. But I think the ##z## with ##h(z)=0##, which is also a root of ##f(x)## cannot be real. This limits the number of possible ##n##, as the graph of ##f(x)## or ##h(x)## must not pass the ##x-##axis. For similar reasons I think (but am not sure yet), that ##r## has to be ##\pm 1##. That would leave us with very special cases for ##h(x)## and therefore also for ##f(x)##.
     
  18. Jan 27, 2017 #17
    there's a solution to this problem, i am just not quite ready to give up just yet. i'll get back to this problem once i have gained more experience and knowledge. i'll make sure that i update this post.
     
  19. Jan 28, 2017 #18

    fresh_42

    Staff: Mentor

    I think I got it. Consider ##a_n := \sqrt[n]{i}\cdot 2^{-\frac{1}{2n}}## and ##f(a_n^n)\cdot q(a_n^n)=f(g(a_n^n))## and next the iteration of ##f(a)=0##.
     
  20. Jan 28, 2017 #19
    please elaborate. i do not understand where that came from and do you have a value for it?
     
  21. Jan 28, 2017 #20

    fresh_42

    Staff: Mentor

    With ##a_n:=\sqrt[n]{i}\cdot 2^{-\frac{1}{2n}}## we get for the integer polynomial ##f(x) \in \mathbb{Z}[x] \subseteq \mathbb{Q}[x] \subseteq \mathbb{C}[x]## with ##f(x)\cdot q(x)=f(g(x))## and ##g(x)=2x^3+x=x \cdot (2x^2+1)##
    $$g(a_n^n)=a_n^n\cdot (2a_n^{2n}+1)=a_n^n \cdot (2(a_n^n)^2+1)= a_n^n \cdot (2\cdot i^2 \cdot \frac{1}{\sqrt{2}^2}+1)=0$$
    and thus ##f(a_n^n)\cdot q(a_n^n)=f(g(a_n^n))=f(0)\,.## If ##f(0)=0## then we have a polynomial of finite degree on the left, that has infinitely many (complex) zeros. We don't have to bother the ##i-##term which might or might not be periodic, because the real factor guarantees that all ##a_n^n## are different. So ##f(x)=0## in this case, which is a possible solution. Hence we may assume ##f(0)\neq 0\,.##

    Now we can consider an arbitrary zero of ##f(x)##, say ##f(a)=0##. By our assumption ##a \neq 0##. The functional equation gives us ##0=f(a)q(a)=f(g(a))## which we can iterate to ##f(g^k(a))=f(\underbrace{g( \ldots (g}_{k-\textrm{times}}(a)\ldots ))=0## with all ##g^k(a) \neq 0##.
    For a polynomial of finite degree, this can only be true if ##g^k(a)=g^l(a)## for some ##k,l## or ##g^n(a)=1## for some ##n \geq 0##.
    So ##g(a)=a^3+a## is a ##n-##th root of unity.

    Now I'm stuck, because I had a typo in my draft (sorry). However, it reduces the number of possibilities a lot, because we know all these roots and the requirement ##f(x)\; , \;f(g(x)) \in \mathbb{Z}[x]## should give further hints.
     
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