Proof that a polynomial has a minimum in $\Bbb R$ I have to prove to following statement and I am having a really hard time here. 
There it is:
Prove that the following polynomial has a minimum in $\Bbb R$
$$p(x)=x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$$
I tried to make the following proof, but I am stuck:
The polynomial can be shown like this:
$$x^4\left(1+\frac{a_3}{x}+\frac{a_2}{x^2}+\frac{a_1}{x^3}+\frac{a_0}{x^4}\right)$$
Let  $$g(x)=\left(1+\frac{a_3}{x}+\frac{a_2}{x^2}+\frac{a_1}{x^3}+\frac{a_0}{x^4}\right)$$
So we can now write the polynomial like
$$p(x)=x^4g(x)$$
Since $x,a \in \Bbb R$ we can say that $\lim _{x \to \infty}=\lim _{x \to -\infty}=1$ (from limit arithmetic).
We know that it is always the case that $x^4>0$, so the sign of $$p(x)=x^4g(x)$$ is determined by $g(x)$. For certain values, $g(x) <0 $ so we can say that there is some $a,b \in \Bbb R$ such that $g(a)<0$ and $g(b)>0$.
$p(x)$ is continouos everywhere, certainly in $[a,b]$, so from the Extreme Value Theorem, the function has a minimum in $[a,b]$.
There are two problems in my proof:


*

*I can't find any value in which $g(x)$ is negative, only if there is some $a$ which is negative. What if $a$ is always positive?

*If the first problem is solved, I managed to prove the statement for some $[a,b]$. Is it just enough? It seems to me that it isn't.
Thanks,
Alan

One could also note that the derivative of an even degree polynomial is an odd degree polynomial.  And an odd degree polynomial always has at least one zero.
