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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
You are copying a lot of problems word-for-word out of textbooks, which is a bad thing for a number of reasons. One, it shows no effort on your part. Two, it gives no indication of what you know, and where we have to start to give a good answer. Three, it's plagiarism to copy something without citing the source. Four, if this is homework, that's OK, but you should add the homework tag. I'm sure there's a fifth and a sixth, but, please, just get with the program. – Gerry Myerson Apr 3 '12 at 7:55

1 Answer 1

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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:


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


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:


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

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