# Let $p(x) = a_0 + a_1 x + .. + a_n x^{n}$. Assume $a_n \neq 0$ and n is odd. Prove $\exists$ x such that p(x)=0? [duplicate]

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!

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.

• I think the solution given by me is 'almost' same,its just that i have asked OP to fill the details! – Arpit Kansal Dec 18 '16 at 20:32
• It's not alike at all in my opinion! I didn't use the Fundamental Theorem of Algebra or anything close to that. – Theorem Dec 18 '16 at 20:33
• @Theorem: I think Arpit Kansal was refering to his solution1. – Math Lover Dec 18 '16 at 20:34
• Sorry I did not see that as he edited that in after I started writing. – Theorem Dec 18 '16 at 20:34
• Dear @Theorem: No problem,i edited just because OP said he don't wanna use FTA! – Arpit Kansal Dec 18 '16 at 20:37

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

• 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. – Nikitau Dec 18 '16 at 20:26
• Dear @Nikitau: i've added one more solution! – Arpit Kansal Dec 18 '16 at 20:28
• Thank you for the hint! :) – Nikitau Dec 18 '16 at 21:20