Show that $p(x) = a_nx^n+a...+a_1x+a_0$ either has a root or attains minimum value in R. Show that if $p(x) = a_nx^n +\dots+ a_1x + a_0$ and $a_n > 0$, then either $p(x) = 0$ has a solution, or else $p(x)$ has attains minimum value on $\mathbb{R}$.
I'm sorry I don't know how to even start the problem. I know that if it is an odd degree polynomial then it has roots. 
Thank you. The second part confuses me the most. This is from an undergrad Real Analysis homework. 

We recall the well-known fact that polynomials are continuous; thus:
If such a polynomial $p(x)$,
$p(x) = \displaystyle \sum_0^n a_i x^i, \tag 1$
with $a_n > 0$, has no roots, then $n = \deg p(x)$ is even; for if $\deg p(x)$ were odd, then since $a_n > 0$, and the highest degree term $a_n x^n$ dominates for sufficiently large $\vert x \vert$, that is,
$x \to -\infty \Longrightarrow p(x) \to -\infty, \tag 2$
$x \to \infty \Longrightarrow p(x) \to \infty; \tag 3$
it is but a short step, via the intermediate value theorem, to see that $p(x)$ has a zero in $\Bbb R$.  So if $p(x)$ has no root, then $\deg p(x)$ is even.  But then we have, again since $a_n > 0$, that 
$\vert x \vert \to \infty \Longrightarrow p(x) \to \infty; \tag 4$
this means that, given any $B > 0$, there exists $L>0$ such that
$\vert x \vert > L \Longrightarrow p(x) > B; \tag 5$
we also observe that
$p(0) = a_0 > 0; \tag 6$
since $p(x)$ never $0$, it cannot change sign; thus (6) follows from (4)-(5).  If we now choose $B > a_0$, and $L$ accordingly large, we affirm that the continuous function $p(x)$ attains its minimum value $p_{min} \le a_0$ at some $x_0 \in [-L, L]$.  
And that is indeed that, is it not?
