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Polynomial Help

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data

    find a polynomial P(x) which has nonnegative coefficients. If P(1)=1 and P(5)= 426, then wast is p(3)

    2. Relevant equations
    P(1)= 6
    P(5)= 426
    P(3)= x

    3. The attempt at a solution
    I have tried to use guess and check. I cant find a way to solve algebraically.
    Last edited: Sep 19, 2010
  2. jcsd
  3. Sep 19, 2010 #2


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    Do you have a specified degree for p(x) ?
  4. Sep 19, 2010 #3
    There is no specified degree for p(X). It just has to have nonnegative coefficents
  5. Sep 19, 2010 #4


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    Then it is best to choose a degree and work with that.
  6. Sep 19, 2010 #5
    There is only one polynomial that works. One hint I received was that it had to be less than 6.
  7. Sep 19, 2010 #6


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    I have no idea how to approach it either. Can't quite gather what to do with the fact that all the coefficients aren't negative. Either way, it must be at least a polynomial of degree 4. But what suggests that it must be less than 6?
  8. Sep 19, 2010 #7
    how do you know it must be at least degree 4?
  9. Sep 19, 2010 #8


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    If you try to fit for example a quadratic

    [tex]ax^2+bx+c[/tex] to the conditions, then you get two conditions:



    We can subtract one from the other to get


    Now we know that a+b+c=1 must be satisfied, with all non-negative numbers, so none of a, b or c can be larger than 1. That's clearly not possible if 24a+4b=425. A similar argument kills cubic polynomials
  10. Sep 19, 2010 #9
    i am afraid i made a mistake when I posted the question. Actually P(1)= 6. So, could you try again. I would appreciate it.
  11. Sep 19, 2010 #10


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    There are lots of choices of polynomials actually, just consider a degree n polynomial of the form axn+b. We know that a+b=6, and that 5na+b=426. Two equations, two unknowns, start trying to find solutions!

    If you know a little linear algebra/convex geometry there's a neat explanation for how you can decide exactly which coefficients are able to be non-zero
    Last edited: Sep 19, 2010
  12. Sep 20, 2010 #11


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    welcome to pf!

    hi chillfactor! welcome to pf! :wink:
    i assume all the coefficients must be whole numbers?

    then ∑an = 6 and ∑an5n = 426

    obviously a0 = 1,

    and by subtracting we have 4a1 + 24a2 + 124a3 = 420 …

    carry on from there :smile:
  13. Sep 20, 2010 #12
    Re: welcome to pf!

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