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By considering large positive and large negative values of $x$ show that the polynomial $a_{2n+1}x^{2n+1} + a_{2n}x^{2n} + ... + a_{1}x + a_{0}$, where $a_{2n+1} \neq 0$, has at least one zero on the real line. Would the same argument show that a polynomial of even degree must have a real zero?

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What did you try? Let $p$ the polynomial: what are $\lim_{x\to -\infty}p(x)$? And $\lim_{x\to +\infty}p(x)$? – Davide Giraudo Apr 3 '12 at 7:47
up vote 1 down vote accepted

Let us denote:

$$P(x)=a_{2n+1}x^{2n+1} + a_{2n}x^{2n} + ... + a_{1}x + a_{0}\ \ ,\ \ a_{2n+1}\neq 0 $$

$P(x)$ as polynomial is continuous function. Let us re-write $P(x)$ as following:

$$P(x)=x^{2n+1}(a_{2n+1}+\frac{a_{2n}}{x}+...+\frac{a_0}{x^{2n+1}})$$

Now, if $|x|$ is big enough then:

$$\text{sgn}(a_{2n+1}+\frac{a_{2n}}{x}+...+\frac{a_0}{x^{2n+1}})=\text{sgn}(a_{2n+1})$$

Therefor, we could find $t>0$ such that, $\text{sgn}(P(t))=\text{sgn}(a_{2n+1})$, and because $2n+1$ is odd we also could find $t'<0$ such that, $\text{sgn}(P(t'))=-\text{sgn}(a_{2n+1})$, hence:

$$P(t)P(t')<0$$

By IVT we conclude that there exist point $c\in (t',t)$ such that $P(c)=0$.

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