Two different roots for $P(x) = x^4+ax^2+bx+c$ 
Let $a, b, c \in \mathbb{R}$ and $a > 0$. Also let $P: \mathbb{R} \to \mathbb{R}$, $P(x) = x^4+ax^2+bx+c$. Show that the function has at most two different roots.

My assumption was to use Bolzano's theorem, but I couldn't figure out how to use it here. Also I'm curious if we could use something like Vieta's formula here? Any help would be appreciated.

Let me try to explain this in detail by simplifying the question as follows: 
Let $a=1>0, |b|=1, |c|=1$. Then we will see 4 different cases and use Descarte's Rule of Signs (*) (http://web.cs.iastate.edu/~cs577/handouts/polyroots.pdf)
$\textbf{Case 1}$: $b=c=1$. Then, $P(x)=x^4+x^2+x+1$ and by (*) it doesn't have any positive roots. However, $P(-x)=x^4+x^2-x+1$ this has 2 sign changes that implies that the function has 2 negative roots. Hence, in total, $2$ real roots.
$\textbf{Case 2}$: $b=1,c=-1$. Then, $P(x)=x^4+x^2+x-1$ and by (*) it has one sign change $\rightarrow$ one positive root. Also, $P(-x)=x^4+x^2-x-1$ this has 1 sign change that implies that the function has 1 negative root1. Hence, in total, $2$ real roots.
$\textbf{Case 3}$: $b=-1,c=1$. Then, $P(x)=x^4+x^2-x+1$ and by (*) it has one sign change $\rightarrow$ one positive root. However, $P(-x)=x^4+x^2+x+1$ this doesn't have any sign changes $\rightarrow$ no negative roots. Hence, in total, $1$ real roots.
$\textbf{Case 4}$: $b=-1,c=-1$. Then, $P(x)=x^4+x^2-x-1$ and by (*) it has one sign change $\rightarrow$ one positive root. Also, $P(-x)=x^4+x^2+x-1$ this has one sign change $\rightarrow$ 1 negative root. Hence, in total, $2$ real roots.
