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Prove that a cubic has no rational roots

  1. Mar 24, 2008 #1
    1) Prove that the acute angle whose cosine is 1/10 cannot be trisected with straightedge and compass.

    ...
    I worked it out and at the end found out that , if I can prove that the cubic polynomial 40x3 - 30x -1 has no rational roots, then I am done.

    Now, is there any way to prove (e.g. divisibility, elementary number theory) that 40x3 - 30x -1 has no rational roots without using the rational root test? (since it is very long and tedious and no calculators are allowed)

    I was trying to prove this by contradiction.
    Suppose x=m/n is a rational root in lowest terms.
    But I am stuck at arriving at a contradiction...how can I possibly do so?


    Any insights?

    Thanks for any help!
     
  2. jcsd
  3. Mar 24, 2008 #2

    tiny-tim

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    Hi kingwinner! :smile:

    The rational root theorem says that any rational root of a_nx^n + … + a_0, can be written as p/q, where p is a factor of a_0 and q is a factor of a_n.

    So any rational root of your equation must be of the form ±1/b, where b is a factor of 40.

    b can only be 1 2 4 8 5 10 20 or 40.

    Now, the turning-points of the function are at 120x^2 = 30, or x = ±1/2.

    At x = 1/2, the function is -11. At x = 0, it is -1.

    So between x = 0 and 1/2 it is negative.

    So that immediately eliminates all possible positive roots except 1.

    At x = -1/2, it is 9. So there is one root between -1/2 and 0.

    If I were you, I'd start by checking one of the highest possible negative roots! :smile:
     
  4. Mar 24, 2008 #3
    Hi,

    Thanks very much! But is there any way to prove it using divisibility and elementary number theory? (since this question is from a non-calculus based course)

    I have a rough idea, but just can't finish it:
    We want to prove that 40x3 - 30x -1 has no rational roots
    40x3 - 30x -1 = 0
    Suppose (proof by contradiction) x=m/n is a rational root in lowest terms
    => 40m3 - 30mn2 - n3 =0
    => 40m3 - 30mn2 = n3

    If p is a prime and p|m, then p|n3 => p|n since p is a prime which contradicts that x=m/n is in lowest terms => no such p => m=+/-1

    Similarly, if q is a prime and q|n, then q|40m3 , and now I am stuck...since this argument above does not seem to work here due to the "40" appearing here...
     
  5. Mar 24, 2008 #4

    tiny-tim

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    Yes, very good! You have just proved half of the rational root theorem! :smile:

    In the letters of my previous post, your m is my p, which has to be ±1.

    And your n is my q = b, which has to be 1 2 4 8 5 10 20 or 40.
     
  6. Mar 25, 2008 #5
    ...Similarly, if q is a prime and q|n, then q|40m3
    =>q|m or q|40 (since q is prime)

    q|m => contradicts gcd(m,n)=1 => no such q => n=+/-1 => x = m/n = +/-1
    q|40=> ?????

    What does q|40 imply?
     
  7. Mar 25, 2008 #6

    tiny-tim

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    Nothing more than ±q = 1 2 4 8 5 10 20 or 40, I'm afraid. :redface:
     
  8. Mar 25, 2008 #7
    Oh, that's right! So I think the rational root test is the only way to go...after all
     
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