Using IVT prove that a polynomial of even degree has atleast two real roots if $a_n a_0 \lt 0$ $P(x) = a_n x^{n} + a_{n-1}x^{n-1} + \cdots +a_1x+ a_0$
$n = 2k$
Show that $P(x)$ has at least two real roots if $a_na_0 \lt 0$

I think I need to find some interval of length $|N|$ in which the graph has a different sign compared to the sign everywhere else. Any ideas/help on how to start  ? Thanks in advance!
 A: Without loss of generality we may assume that $a_n\gt 0$. For if it is not, we can multiply $P(x)$ by $-1$ wiithout changing the roots. Then $a_0\lt 0$.
Note that $P(0)\lt 0$. If we can show that $P(a)\gt 0$ for some positive $a$, it will follow by the Intermediate Value Theorem that $P(x)=0$ for some $x$ between $0$ and $a$, that is, for some positive $x$.
By dividing top and bottom of 
$$\frac{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0}{a_nx^n}\tag{1}$$ by $a_nx^n$ we can show that
$$\lim_{x\to\infty} \frac{P(x)}{a_nx^n}=1.\tag{1}$$
In particular, if $x$ is large enough positive, $P(x)$ is positive, so there is a positive $a$ such that $P(a)\gt 0$.
We now show that there is a negative number $-b$ such that $P(-b)\gt 0$. That will imply that $P(x)=0$ has a root between $-b$ and $0$, that is, a negative root.
An argument essentially identical to the argument we used for (1) shows that
$$\lim_{x\to-\infty} \frac{P(x)}{a_nx^n}=1.\tag{2}$$
Since $n$ is even, for negative $x$ we have that $a_nx^n$ is positive. So if $x$ is large enough negative, by (2) $P(x)$ is positive. This completes the proof.
A: Suppose $a_0 < 0$ and $a_n >0$
Firstly we have $P(0) = a_0 < 0$.
$P$ is of even degree, which means $\lim_{x \to +\infty} P(x) = \lim_{x\to -\infty}P(x) = +\infty$, since $a_n > 0$
Apply IVT on $[0, +\infty)$ and $(-\infty, 0]$, we find two roots for $P(x)$.
If $a_0 > 0$ and $a_n <0$ we can argue in a similar way.
A: Hints
The polynomial is of even degree.
What does this means about the limits at $x\rightarrow-\infty$ and $+\infty$?
Now what is $P(0)$? Using the fact that $a_na_0<0$ means that $a_n$ and $a_0$ are of opposite signs, you should be able to finish.
