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Polynomial inequality

  1. Oct 25, 2007 #1

    Gib Z

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    [SOLVED] Polynomial inequality

    1. The problem statement, all variables and given/known data

    Show [tex]x^4-3x^3+4x^2-3x+1 \geq 0[/tex] for all real x.

    2. Relevant equations
    [tex]a^2 \geq 0[/tex]
    Cant think of any more.


    3. The attempt at a solution

    I've tried to factor it relentlessly, because I would have sworn on my life there was a nice trick for factoring polynomials with symmetrical and alternating co efficients, but I cant remember...By observation i can see the inequality is trivial if x is less than 0 or greater than 3. But I can't fill the gap.
     
  2. jcsd
  3. Oct 25, 2007 #2

    Dick

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    Forget the tricks. If a is a root then (x-a) is a factor. Look for obvious roots. There's an easy way to list all possible rational roots as well, right?
     
  4. Oct 25, 2007 #3
    Think about (x - 1)^3 = x^3 - 3x^2 + 3X -1

    When you have A(x).B(x) >= 0, and you know A(x) >= 0 for all x then you prove B(x) >= 0 for all real x.
     
  5. Oct 25, 2007 #4
    (x-1)^2 = x^2 - 2x + 1
    (x-1)^3 = x^3 - 3x^2 + 3x - 1
    (x-1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1
    ...
     
  6. Oct 25, 2007 #5

    dynamicsolo

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    Dick's suggestion is the most straightforward for this one. Have you had the Rational Zeroes Theorem? One of the candidates from that technique works rather easily. From there, you'll be able to get all the zeroes. Factorization of this polynomial plus what multiplicity of zeroes does to the x-intercepts will tell you most of what you need to know.
     
  7. Oct 25, 2007 #6
    The question gets easier as you go along. I'll do this completely as I want to show a way of 'dividing' polynomials that is much quicker.

    So it is clear that x=1 is a root by calculation. Hence (x-1)f(x) is our desired polynomial where the degree of f(x) is strictly less than 4. It is clear that the coefficient on the third power must be 1 as (x-1)f(x) must yield a 4th degree polynomial. Similarly the constant term must be -1, as (x-1)f(x) must yield a polynomial with constant term 1.

    So what is the coefficient for the second power? Suppose it was a, then coefficient of the third power of (x-1)f(x) will be (1)a+(-1)1=-3. So a=-2. So the coefficient for the first power becomes 2 by similar arguments.

    So we have (x-1)(x^3-2x^2+2x-1). We can break up the latter product again knowing it has a root of 1. By the same arguments we get (x^3-2x^2+2x-1)=(x-1)(x^2-x+1).

    Combining our equation is (x-1)^2(x^2-x+1) and completing the square gives (x-1)^2((x-1/2)^2+3/4) which is clearly nonnegative.

    This technique was never taught to me at school. I don't know how many people catch on to this; it was just one of the things I've always done for fast computations.
     
  8. Oct 25, 2007 #7

    cristo

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    Your method is just dividing one polynomial into another (i.e. divide (x-1) into the original polynomial). Instead of writing it down in the long division type way, you've just written it down in words, but it's basically exactly the same method.
     
  9. Oct 25, 2007 #8
    I realize this. But the point is you can do it all in your head. Anyways, that's why I called it 'division'.
     
  10. Oct 25, 2007 #9

    dynamicsolo

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    The completion of the square is really what cinches the proof. (And spares one the nuisance of having to show that x^2 - x >= -1 .)

    Once again, though, we've been terribly nice and solved the problem. I wonder if the OP has even seen any of this yet...
     
  11. Oct 25, 2007 #10

    Dick

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    The OP is a Homework Helper. I suspect he's just on vacation or something. I'm not worried about him.
     
  12. Oct 25, 2007 #11

    dynamicsolo

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    Sorry, I missed that... (I've seen threads where the helpers got a bit too enthusiastic, so I've been making an effort to restrain myself from saying too much -- unlike one of my regular jobs where I pretty much walk students through problems.)
     
  13. Oct 25, 2007 #12

    Dick

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    You don't have to apologize. Jeez. I just thought it was funny.
     
  14. Oct 26, 2007 #13

    Gib Z

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    Hey guys I'm not on Vacation, I just couldn't get on the computer yesterday because my mother had some relatives over. I think you guys did in fact go a little over the top in helping =], Dick's first post jolted my brain awake :) And it also reminded me of the "trick" that I was thinking of, though I remember it only works when there is an even number of terms. When the coefficients are symmetrical and alternating in signs, 1 will be a root because the polynomial will be reduced to the sum of its coefficients, which will be zero. That works in this case only after the first division by (x-1).

    Anyway, Yes I reduced it to the perfect square and irreducible quadratic, Thank you guys very much =]
     
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