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I have known a theorem from the link Derivation of the general forms of partial fractions

The theorem is that a polynomial of odd degree has a root.But I don't know its proof.Can anyone prove it with intuition.

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Shahed al mamun
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Suppose the polynomial $P$ is monic, for this is irrelevant as the existence of a zero. Also, I presume, it's over the reals, for else this is just not true.

Then $\lim_{x\to-\infty}P(x)=-\infty$ and $\lim_{x\to\infty}P(x)=\infty$ because the polynomial is dominated by its leading term, which is an odd function -- it's $x^{2n+1}$. Since $P$ is a continuous function with both a positive and negative value, it must have a zero. Pictorially, if it has a value above and below the $x$-axis and the graph is continuous, then it has to cross the $x$-axis.

I am not entirely sure why you refer to partial fractions.

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r0fg1
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  • I mentioned partial fractions because I'm trying to learn the derivation of the methods of fraction decomposition.And the above theorem is required for this purpose. – Shahed al mamun Feb 21 '15 at 05:20