**Theorem** Let $P: \mathbb{R} \to \mathbb{R}$ be given by $P(x)=\sum_{k=0}^{2n} a_{k}x^{k}, \quad a_{2n}>0,$(i.e. an *even degree polynomial* with positive leading coefficient). Then $P$ has a minimum at some point $x=x_{*}$ and 1. if $P(x_{*})>0$ then there are no real roots 2. if $P(x_{*}) = 0$ then there is at least one real roots 3. if $P(x_{*})<0$ then there are at least two real roots **Proof** For $|x|>1$, we have $\begin{align} P(x) &\geq a_{2n}|x|^{2n} - \sum_{k=0}^{2n-1} |a_{k}||x|^{k} \\ & \geq a_{2n} |x|^{2n} - \left( \sum_{k=0}^{2n-1} |a_{k}| \right) |x|^{2n-1} \end{align}$So $\exists R\geq1$ s.t. $P(x)\geq \max(P(0)+1,1)$ for all $|x| \geq R$ Applying [[Extreme Value Theorem|EVT]] to $f:[-R,R]\to \mathbb{R}$ gives that $\exists x_{*} \in [-R,R]$ such that $P(x)\geq P(x_{*}) \quad \forall x\in[-R,R].$ Then for $|x|\geq R$ we have $P(x)\geq P(0)\geq P(x_{*})$ since $0\in[-R,R]$. Therefore $P(x) \geq P(x_{*})$ for all $x \in \mathbb{R}$ and we have found the minimum of $P$. Parts (1) and (2) are immediate. Since $P(-R)\geq 1>0$, $P(x_{*})lt;0, & $P(R) \geq 1>0$, part (3) follows using [[Intermediate Value Theorem (IVT)|IVT]] on the intervals $[-R,x_{*}]$ and $[x_{*},R]$.