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Math Help - polynomial root

  1. #1
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    polynomial root

    If we are told that a+\frac{b}{2}+\frac{c}{3}+\frac{d}{5}=-\frac{e}{6}, where a,b,c,d,e \in {R}.

    Prove that the polynomial f(x)=a+bx+cx^2+dx^4+ex^5 has at least one real zero.
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  2. #2
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    Quote Originally Posted by putnam120 View Post
    If we are told that a+\frac{b}{2}+\frac{c}{3}+\frac{d}{5}=-\frac{e}{6}, where a,b,c,d,e \in {R}.

    Prove that the polynomial f(x)=a+bx+cx^2+dx^4+ex^5 has at least one real zero.
    As:

    x \to \infty f(x) \to + \infty

    and:

    x \to -\infty f(x) \to - \infty

    f(x) as its continuous has a real root.

    In fact all odd order polynomial with real coefficients must have at
    least one real root.

    RonL
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  3. #3
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    Quote Originally Posted by putnam120 View Post
    If we are told that a+\frac{b}{2}+\frac{c}{3}+\frac{d}{5}=-\frac{e}{6}, where a,b,c,d,e \in {R}.

    Prove that the polynomial f(x)=a+bx+cx^2+dx^4+ex^5 has at least one real zero.
    Use mean value theorem for integral.
    Consider the continous function on [0,1], f(x)=a+bx+cx^2+dx^4+ex^5.

    By the integral mean value theorem there is a number c\in [0,1] such as,
    f(c)(1-0)=\int_0^1 a+bx+cx^2+dx^4+ex^5dx
    But that tells us that,
    f(c)=ax+\frac{bx^2}{2}+\frac{cx^3}{3}+\frac{dx^5}{  5}+\frac{ex^6}{6} \big|_0^1
    Evaluate,
    f(c)=a+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}+\frac{e  }{6}=0
    Note only we know it has a solution we know it has a solution on the interval [0,1]
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  4. #4
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    Quote Originally Posted by CaptainBlank View Post

    In fact all odd order polynomial with real coefficients must have at
    least one real root.
    But how do you know that e\not = 0
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    But how do you know that e\not = 0
    Well this partiular statement is independent of e as I said any odd order
    polynomial, but in this case yes I suppose e could in principle
    be zero.

    In fact I do have another method using Descartes rule of signs which
    does use the condition, which I beleive handles the possibility that
    e=0 (but I don't intend to check that now).

    RonL
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