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If $r(x)$ is an $n$th degree polynomial of the form $r(x) = a_0+a_1x+\cdots+a_nx^n$ with $n$ odd, prove that $r(x)$ has at least one real root. In what case does equality hold?

Attempt:

We have that $r(x) = a_nx^n+(a_{n-1}x^{n-1}+\cdots+a_1x+a_0) = a_nx^n + O(x^{n-1})$ as $x \to \pm \infty$. Thus there exists a point $x_0$ such that $r(x_0) > 0$ and a point $x_1$ such that $r(x_1)<0$. Therefore since $r(x)$ is continuous, it follows by the intermediate value theorem that $r(x)$ has at least one real root.

Equality holds when a real root has odd multiplicity and the rest of the roots are imaginary. That is, $$r(x) = (x-a)^{2m+1}Q(x)$$ where $Q(x)$ is an $n-2m-1$ degree polynomial with all roots imaginary for some $a$ and $m$.

Questions: Does my proof look good and also for the case of equality is there a way to simplify it even more than that?

Puzzled417
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  • You should at least state that $r(x)\to\operatorname{sign}\pm\infty$ as $x\to\infty$, where the sign depends on the sign of $a_n$, and similarly for $x\to-\infty$, and then conclude that there exists points $x_0$ and $x_1$ as you stated. I don't agree with using notation of order $O(x^{n-1})$ in a proof at this level (at most, just for some drafts), but if that is how you (and your professor) do in class then it is how you should do it. You also didn't explain what equality should hold or not!? – Luiz Cordeiro May 03 '16 at 03:43
  • Why does equality hold in these situations? That condition may be true, but you haven't motivated it.

    Your proof for the first part looks good.

     – Mark Schultz-Wu May 03 '16 at 03:44
  • @Mark It follows from the complex conjugate root theorem. – Puzzled417 May 03 '16 at 03:45

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