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Polynomial Question

  1. Nov 9, 2004 #1

    uart

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    Somebody gave me the following question, I was able to solve it but was unsure about some of the assumptions involved.

    Question : Given that the polynomial x^3 + m x^2 + 15 x - 7 has at least two rational roots then find m.

    Now the question didn't state that m had to be integer and I was unsure as to whether this was meant to be assumed or whether it could be deduced.

    Here's what I did.

    1. Since the product of the roots is 7 then two rational roots implies that the third root is also rational.

    2. I assumed that m was integer which meant that the rational roots where also integer. ( by the http://planetmath.org/encyclopedia/RationalRootTheorem.html [Broken] )

    3. Since there are very few ways of having integer roots that multiply to give 7 I easily found the possible roots of 7, 1, 1, that multiply to give 7 and also have sum of pair-products totalling to 15.

    4. So m = -(7 + 1 + 1) = -9


    So did I need to assume that m was integer or could it have been deduced?
     
    Last edited by a moderator: May 1, 2017
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  3. Nov 9, 2004 #2

    matt grime

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    You've found an m that satisfies the requirements. What does it matter that you assumed m was integer? It makes it easier and gives a unique solution, but so what? Had the assumption not yielded an answer then you couldn't conclude no solution existed, but that's all.

    7=xyz
    15=xy+xz+yz

    15=xy+x(7/xy)+y(7/xy)=xy+7y+7x

    so let x=2,

    15=2y+7y+14, so y=1/9, and hence z=63/2 would also work, so it is neither necessary for m to be an integer, nor is it deducible that m is one.
     
    Last edited: Nov 9, 2004
  4. Nov 9, 2004 #3

    uart

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    Yes I know that the solution is ok, and I'm sure it was the intended solution, but I am still uncertain as to whether it's unique.

    Is there any theroem that would have implied that m had to be integer once I had all three roots rational, all other coefficients integer, and a monic polynomial ?
     
  5. Nov 9, 2004 #4

    matt grime

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    See my example, in the above edited post. Clearly it is not necessary for m to be an integer, and cannot be deduced.
     
  6. Nov 9, 2004 #5

    uart

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    Thanks Matt, that's just what I wanted to know. :)
     
  7. Nov 9, 2004 #6

    uart

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    Hang on Matt, there is a mistake in that example.

    It should be 15 = xy + 7/x + 7/y.

    x=2 leads to a quadratic with irrational solutions for y.
     
    Last edited: Nov 9, 2004
  8. Nov 9, 2004 #7

    matt grime

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    It's also clear that the choice of x=2 was completely arbitrary, and any rational would have yielded an answer.
     
  9. Nov 9, 2004 #8

    uart

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    With the corrected equation (15 = xy + 7/x + 7/y), I don't think there is any possible rational solution for x and y other than the integer solution (x=1, y=1). I'm not sure how to prove that though.
     
  10. Nov 9, 2004 #9

    matt grime

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    Damn, buggered that one up then.

    need 15xy=x^2y^2 +7y+7x, so y is rational iff (7-15x)^2 - 28x^3 has rational square root (for some rational x) which seems too hard to solve, so I think we should assume that they meant that m was an integer, but let's try: we need to find a/b=x such that

    (7b-15a)^2/b^2 - 28a^3/b^3 is a perfect square, If we let b be a square number, then it suffices (though is not necessary) that

    b(7b-15a)^2-28a^3 is also a square. Anyone any suggestions?

    In general though there is no theorem that states: if P is a monic poly with rational roots, then all but one coeff integer implies the last one must be: consider

    x^2 - ax+1, If a=2, then there are two integer solutions, though in general

    a=p+q where p=1/q and p is any rational will do, and I don't see it changing generally for higher degree polys: it really will depend on the poly in question.
     
    Last edited: Nov 9, 2004
  11. Nov 9, 2004 #10

    JasonRox

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    What if m=-x? No one said it was a constant.
     
  12. Nov 9, 2004 #11
    It was stated that it is a polynomial. And, polynomials are in the form of [tex] a_{n}x^{n} + a_{n-1}x^{n-1} + a_{n-2}x^{n-2}+...+a_{2}x^{2}+a_{1}x^{1}+a_{0}[/tex] where the a's are constant coefficients. So it is assumed that "m" in his problem would be a constant coefficient.
     
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