Proof that a polynomial has a minimum in 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 R

$$p(x)={x}^{4}+{a}_{3}{x}^{3}+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$$

I tried to make the following proof, but I am stuck:

The polynomial can be shown like this:

$${x}^{4}(1+\frac{{a}_{3}}{x}+\frac{{a}_{2}}{{x}^{2}}+\frac{{a}_{1}}{{x}^{3}}+\frac{{a}_{0}}{{x}^{4}})$$

Let

$$g(x)=(1+\frac{{a}_{3}}{x}+\frac{{a}_{2}}{{x}^{2}}+\frac{{a}_{1}}{{x}^{3}}+\frac{{a}_{0}}{{x}^{4}})$$

So we can now write the polynomial like

$$p(x)={x}^{4}g(x)$$

Since $x,a\in \mathbb{R}$ we can say that $\underset{x\to \mathrm{\infty}}{lim}=\underset{x\to -\mathrm{\infty}}{lim}=1$ (from limit arithmetic).

We know that it is always the case that ${x}^{4}>0$, so the sign of

$$p(x)={x}^{4}g(x)$$

is determined by g(x). For certain values, $g(x)<0$ so we can say that there is some $a,b\in \mathbb{R}$ 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.

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 R

$$p(x)={x}^{4}+{a}_{3}{x}^{3}+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$$

I tried to make the following proof, but I am stuck:

The polynomial can be shown like this:

$${x}^{4}(1+\frac{{a}_{3}}{x}+\frac{{a}_{2}}{{x}^{2}}+\frac{{a}_{1}}{{x}^{3}}+\frac{{a}_{0}}{{x}^{4}})$$

Let

$$g(x)=(1+\frac{{a}_{3}}{x}+\frac{{a}_{2}}{{x}^{2}}+\frac{{a}_{1}}{{x}^{3}}+\frac{{a}_{0}}{{x}^{4}})$$

So we can now write the polynomial like

$$p(x)={x}^{4}g(x)$$

Since $x,a\in \mathbb{R}$ we can say that $\underset{x\to \mathrm{\infty}}{lim}=\underset{x\to -\mathrm{\infty}}{lim}=1$ (from limit arithmetic).

We know that it is always the case that ${x}^{4}>0$, so the sign of

$$p(x)={x}^{4}g(x)$$

is determined by g(x). For certain values, $g(x)<0$ so we can say that there is some $a,b\in \mathbb{R}$ 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.