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I have no idea how to solve this question. I'm trying to show that given $p(x) = a_0 + a_1 x + .. + a_n x^{n}$ and further assuming that $a_n \neq 0$ and n is odd, there exists x such that p(x)=0.

I'm guessing that Rolle's theorem might come into play somewhere, since I know that by Rolle's theorem if $f:[a,b] \rightarrow \mathbb{R}$ is continuous on [a,b] and differentiable on (a,b) and f(a) = f(b) then $\exists c \in (ab)$ s.t f'(c)=0. Other than this hunch, I have no idea where to begin solving this.

Help would be much appreciated!

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Assuming you want to use calculus rather than linear algebra, use the Intermediate value theorem. Without loss of generality, assume $a_n=1$ (why is this allowed?). Then

$P(x)=x^n(1+\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots +\frac{a_{0}}{x^n})=x^n g(x)$ for $x\neq0$. Notice how $\lim_{x\rightarrow\pm\infty} g(x)=1$ so for large enough $x$'s, $g(x)>0$ and hence $P$'s sign is determined by $x^n$ for large $x's$.

It follows there are $a<0,b>0$ such that $P(a)=a^n g(a)<0,\ P(b)=b^ng(b)>0$. Use $P$'s continuity to conclude there's a $c$ such that $P(c)=0$

This is the idea. If needed more elaboration don't hesitate to ask.

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    $\begingroup$ I think the solution given by me is 'almost' same,its just that i have asked OP to fill the details! $\endgroup$ – Arpit Kansal Dec 18 '16 at 20:32
  • $\begingroup$ It's not alike at all in my opinion! I didn't use the Fundamental Theorem of Algebra or anything close to that. $\endgroup$ – Theorem Dec 18 '16 at 20:33
  • $\begingroup$ @Theorem: I think Arpit Kansal was refering to his solution1. $\endgroup$ – Math Lover Dec 18 '16 at 20:34
  • $\begingroup$ Sorry I did not see that as he edited that in after I started writing. $\endgroup$ – Theorem Dec 18 '16 at 20:34
  • $\begingroup$ Dear @Theorem: No problem,i edited just because OP said he don't wanna use FTA! $\endgroup$ – Arpit Kansal Dec 18 '16 at 20:37
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Here are some steps (without full details) which will lead you to the solution:

Solution 1: WLOG, assume $a_n>0$,now as $x \to \infty$ note that $p(x) \to \infty$ and as $x \to -\infty$ then $p(x) \to -\infty$.Now apply the IVP.

Solution2:

Step 1: Show that $P(\overline z)=\overline{P(z)}$,hence conclude that $z$ is a root of $p(z)$ iff $\overline z$ is a root of $p(z)$

Step 2: Recall Fundamental Theorem of Algebra

Step: Conclude that $p$ has at least one real root. QED

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  • $\begingroup$ unfortunately I've not covered the Fundamental Theorem of Algebra in class. Apparently upon further probing, there was hint stating to use the intermediate value theorem. $\endgroup$ – Nikitau Dec 18 '16 at 20:26
  • $\begingroup$ Dear @Nikitau: i've added one more solution! $\endgroup$ – Arpit Kansal Dec 18 '16 at 20:28
  • $\begingroup$ Thank you for the hint! :) $\endgroup$ – Nikitau Dec 18 '16 at 21:20

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