$\color{crimson}{\text{Problem}}$

Show that if $f:\mathbb{R}\to\mathbb{R}$ be a polynomial of odd degree with real coefficients then it has at least one real root.

$\color{green}{\text{Proof}}$

Let $$f(x)=a\_{2n+1}x^{2n+1}+a\_{2n}x^{2n}+\ldots+a\_1x+a\_0$$ be a polynomial where $a\_i\in\mathbb{R}$ for all $i\in \{1,2,\ldots,2n+1\}$ and $a\_{2n+1}\ne 0$. Without loss of generality we can also assume that $a\_0>0$. Then, $$\boxed{f(0)>0}$$

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> **Case 1**
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Let $a\_i\ne 0$ for all $i\in \{1,2,\ldots,2n+1\}$. Then we rewrite $f(x)$ as,

$$f(x)=x^{2n}\left(a\_{2n+1}x+a\_{2n}\right)+\ldots+(a\_1x+a\_0)$$Now observe that the sign of the terms of the form $x^{2i}\left(a\_{2i+1}x+a\_{2i}\right)$ depends only on the sign of $\left(a\_{2i+1}x+a\_{2i}\right)$. So, if we want to show that $f(x)<0$ for some $x\in\mathbb{R}$, it suffices to show that each terms of the form $\left(a\_{2i+1}x+a\_{2i}\right)<0$. This arguments shows that $f(x\_0)<0$ for any $x\_0\in \mathbb{R}$ such that, $$x\_0<\min\left(-\dfrac{a\_{2n}}{a\_{2n+1}},-\dfrac{a\_{2n-2}}{a\_{2n-1}},\ldots,-\dfrac{a\_{0}}{a\_{1}}\right)$$

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> **Case 2**
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If on the other hand $a\_{2i+1}=0$ for some $i$'s (if $a\_{2i}$'s are $0$, it won't affect the argument of **Case 1**) then we define the polynomial $g:\mathbb{R}\to\mathbb{R}$ defined by, $$g(x)=b\_{2n+1}x^{2n+1}+b\_{2n}x^{2n}+\ldots+b\_1x+b\_0$$ where, $$b\_{2i+1}\begin{cases}=-1 & \text{if}\ a\_{2i+1}=0\\=a\_{2i+1} & \text{if}\ a\_{2i+1}\ne0\end{cases}$$Then, to the polynomial $g(x)$, the argument of **Case 1** applies. 

Now for this $g(x)$ we choose $x\_0$ is such a manner that $x\_0<0$. Then $g(x\_0)<0$ and also $g(x\_0)>f(x\_0)$. Consequently $f(x\_0)<0$

What is left now is to apply [Bolzano's Theorem](https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CB0QFjAAahUKEwj-ubCv8LHHAhVBGo4KHZz8BNY&url=http%3A%2F%2Fmathworld.wolfram.com%2FBolzanosTheorem.html&ei=uL_SVf79GsG0uASc-ZOwDQ&usg=AFQjCNF-cguShyO7lCJyVCdX_2D5uJpG6g) to conclude the existence of a root.