Proving that a polynomial has a positive root So I want to prove that a polynomial $ P(x)=a_nx^n+a_{n−1}x^{n−1}+.....+a_1x+a_0 $ has a positive root. I'm given that $ a_n $ is positive and $ a_0 $ is negative. I want to know how to apply the intermediate value theorem EXACTLY.
Here is what the theorem says:
Let $f$ be a continuous real function on the interval $[a,b]$. If $f(a) \lt f(b)$ and if $c$ is a number such that $f(a) \lt c \lt f(b)$, then there exists a point $x$ in $(a,b)$ such that $f(x)=c$. 
So I am assuming that polynomials are continuous.  given that $P(0) \lt 0 \lt P(x)$ (for $x$ large enough). So there exists a point $c$ such that $f(c)=0$.
I have some gaps in my understanding of the argument..
 A: "So there exists a point $c$ such that $f(c)=0$." That should be "There exists a point $y\in(0,x)$ such that $f(y)=c$." Afterall $c$ in the range value of $f$, and you want to find an $y$ for $c=0$. 
A: Note that since $a_n \gt 0$ and $a_0 \lt 0$, it must be the case that $n \gt 0$ to avoid an inconsistency.
While it is clear that $P(0) = a_0 \lt 0$, the intuitive notion that $P(x)$ attains positive values for all sufficiently large $x$ deserves a little justification.
What we have to work with is that the leading coefficient $a_n \gt 0$ is positive.  Then for any positive $x \gt 0$ we can write:
$$ P(x) = a_n x^n (1 + b_{n-1}x^{-1} + \ldots + b_0 x^{-n}) $$
where $b_i = a_i/a_n$ for $i = 0,\ldots,n-1$.
Thus the goal of getting $x$ large enough to make $P(x) \gt 0$ can be achieved if and only if:
$$ 1 + b_{n-1}x^{-1} + \ldots + b_0 x^{-n} \gt 0 $$
Equivalently, since $x \gt 0$, what we need to achieve is:
$$ 1 \gt \frac{b_{n-1} + b_{n-2}x^{-1} + \ldots + b_0 x^{-n+1}}{x} $$
$$ x \gt b_{n-1} + b_{n-2}x^{-1} + \ldots + b_0 x^{-n+1} $$
An explicit recipe for $x$ that accomplishes this is to take $B = \max\{ |b_i| : i=0,\ldots,n-1\}$ and choose $x$ that is greater than both $1$ and $nB$.
The first fact, $x \gt 1$, implies the reciprocal powers of $x$ in the expression are each less than $1$.  Thus:
$$ nB \ge |b_{n-1}| + |b_{n-2}| + \ldots + |b_0| \gt b_{n-1} + b_{n-2}x^{-1} + \ldots + b_0 x^{-n+1} $$
Then the second fact, $x \gt nB$, finishes the proof of inequality:
$$ x \gt nB \gt b_{n-1} + b_{n-2}x^{-1} + \ldots + b_0 x^{-n+1} $$
In this way every $x$ which is greater than the maximum of $1$ and $nB$ gives $P(x) \gt 0$.
