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Math Help - polynomial question

  1. #1
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    polynomial question

    Hi So I have to prove the following:
    If p(x)=a_{n}x^n+...+a_{1}x+a_{0} with a_{0}<0 and a_{n}>0 and n is an even positive integer, then prove that p has at least two distinct roots.

    So my proof looks a little like this so far:
    p is continuous on \mathbb{R}.
    \lim_{x\to +\infty}p(x)= +\infty and \lim_{x\to -\infty}p(x)= -\infty.
    Hence \exists \alpha, \beta \in \mathbb{R} such that:
    p(\alpha)<0<p(\beta).
    ...
    At this point Im stuck. I found, via graphing example polynomials, that the negative a_{0} values cause the polynomial to have two distinct roots, but how to state this mathematically to finish the proof?
    Any suggestions would be appreciated thanks.
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  2. #2
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    In your conditions the result is not true, just consider p(x)=x^3-1

    If n is odd or n even and a_n<0, p(0)<0, \lim_{x\to-\infty}p(x)=\lim_{x\to\infty}p(x)=\infty.Thus there exists \alpha<0,\beta>0 such that
    p(0)<0<p(\alpha), and p(0)<0<p(\beta).

    Now you only need Bolzano.
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  3. #3
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    the hypothesis states that n is even,
    p(x) = x^3 - 1 does not satisfy the hypothesis
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  4. #4
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    You are right!!!

    I was translating odd an even to spanish strangely!

    OK, if n is even, then with your hypothesis \lim_{x\to-\infty}p(x)=\lim_{x\to-\infty}p(x)=+\infty thus the rest of the argument given in my post works!
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  5. #5
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    Quote Originally Posted by dannyboycurtis View Post
    Hi So I have to prove the following:
    If p(x)=a_{n}x^n+...+a_{1}x+a_{0} with a_{0}<0 and a_{n}>0 and n is an even positive integer, then prove that p has at least two distinct roots.

    So my proof looks a little like this so far:
    p is continuous on \mathbb{R}.
    \lim_{x\to +\infty}p(x)= +\infty and \lim_{x\to -\infty}p(x)= -\infty.
    Hence \exists \alpha, \beta \in \mathbb{R} such that:
    p(\alpha)<0<p(\beta).
    ...
    At this point Im stuck. I found, via graphing example polynomials, that the negative a_{0} values cause the polynomial to have two distinct roots, but how to state this mathematically to finish the proof?
    Any suggestions would be appreciated thanks.
    Use the intermediate value property: If f(x) is a continous function on [a,b], then it takes on all values between f(a) and f(b).

    f(0)< 0 while, for some large x, f(x)> 0.

    f(0)< 0 while for some larger negative x, f(x)> 0.
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  6. #6
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    how could I show that if n was odd, p would NOT have an odd number of distinct roots?
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  7. #7
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    You wouldn't. It is not true. For example, f(x)= x^3- x has odd degree and an odd number of distinct roots.

    If you mean "if n is odd, p may not have an odd number of distinct roots", use the counter example p(x)= x^3.
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  8. #8
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    sorry I miswrote that question,
    how would I prove that if n was odd, that p would not have 2 distinct roots?
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  9. #9
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    Again, you don't prove it. It's not true. For example, x^3- x= 0 has roots x= 0, x= 1, x= -1, three distinct roots, so it would certainly also be correct to say it has 2 distinct roots.

    If you mean "does not have exactly two distinct roots" it is still not true. x^3- 2x^2+ x= 0 has exactly x= 0 and x= 1 as distinct roots.

    What is true is that a polynomial, of odd degree, with real coefficients cannot have exactly two (or any even number) real roots (distinct or not). But that is not anything like what you have been saying.
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