Prove that: if a polynomial of an even degree takes any value which sign is opposite to its highest power coefficient sign, then it has at least two roots in $\Bbb R$.
To prove this statement I've started by formalizing the above. Let: $$ f(x) = a_{2n}x^{2n} + a_{2n-1}x^{2n-1} + \cdots + a_{1}x + a_0 $$
We want to show that there exist at least two points $x', x''$, such that: $$ f(x') = f(x'') = 0 $$
Rewrite the polynomial in the following way: $$ f(x) = a_{2n}x^{2n}\left(1 + {a_{2n-1}\over a_{2n}}\cdot{1\over x} + {a_{2n-2}\over a_{2n}}\cdot{1\over x^2} + \cdots + {a_{1}\over a_{2n}}\cdot{1\over x^{2n-1}} + {a_0\over a_{2n}}\cdot{1\over x^{2n}}\right) $$
Consider the case when $a_{2n} > 0$. Taking the limit as $x\to\infty$ we observe: $$ \lim_{x\to+\infty} f(x) = +\infty\\ \lim_{x\to-\infty} f(x) = +\infty $$
From the problem statement, we know that there exists at least one $x_0 \in\Bbb R$ such that $f(x_0) < 0$ for $a_{2n} > 0$. Now we split the domain of the function into two parts: $x \in (-\infty; x_0)$ and $x\in (x_0; +\infty)$. Now we are going to apply the intermediate value theorem to both intervals.
Since $f(x) \stackrel{x\to-\infty}{\to} +\infty$ and $f(x_0) < 0$, then there must exist $x' \in (-\infty; x_0)$ such that: $$ f(x') = 0 $$
Since $f(x) \stackrel{x\to+\infty}{\to} +\infty$ and $f(x_0) < 0$, then there must exist $x'' \in (x_0; +\infty)$ such that: $$ f(x'') = 0 $$
Thus we have found $x', x''$ such that $f(x) = 0$ at that points. In case of $a_{2n} < 0$ we could either multiply the polynomial by $-1$ and apply the above, or use similar reasoning for negative $a_{2n}$
I would like to ask for the verification of my prove and/or point to the mistakes. Thank you!
This looks good. Just one comment. It's possible that the statement of the Intermediate Value Theorem you are invoking only applies to bounded intervals. In this case, you cannot invoke the theorem on the intervals $(-\infty,x_{0})$ and $(x_{0},+\infty)$. However, since you know $f(x)\to+\infty$ as $x\to-\infty$ and as $x\to+\infty$, you know that there are values $a<x_{0}<b$ such that $f(a),f(b)>0$. Then, you can apply the Intermediate Value Theorem to the bounded intervals $[a,x_{0}]$ and $[x_{0},b]$.