> [!NOTE] Theorem (Fundamental Theorem of Algebra) > Let $p \in \mathbb{C}[x]$ be a non-constant polynomial: $\deg(p)\geq 1$; where $\mathbb{C}[x]$ be the [[Ring of Polynomial Forms|ring of polynomial forms]] in $x$ over the [[Complex Numbers|complex numbers]]. Then there exists $\alpha \in \mathbb{C}$ such that $p(\alpha)=0$. ###### Proof \[Analysis 3\] We will prove the result by contradiction. Assume that $|p(z)| \neq 0$ for ever $z \in \mathbb{C}$. Define $f: \mathbb{C} \rightarrow \mathbb{C}$ by $f(z)=\frac{1}{p(z)}$. Now, since $p$ does not vanish, the function $f$ is [[Complex Differentiability|analytic]] in all of $\mathbb{C}$, since it is the composition of two holomorphic functions ($1 / z$ is holomorphic outside the origin). Notice that if we assume $p(z)=\sum_{k=0}^n c_k z^k$, with $c_n \neq 0(n>0)$, then at infinity the polynomial behaves like $c_n z^n$, as that is the highest power. That means $|p(z)|$ goes to infinity as $z$ goes to infinity, and satisfies $|p(z)|>1$ for all $z \mid>R$ for some $R>0$. As a result the function $f(z)=\frac{1}{p(z)}$ is bounded in $\mathbb{C}$. It is less than $1$ for all $|z|>R$ based on our analysis of $p$, and it is bounded on the compact set $|z| \leqslant R$ since it is continuous. [[Louiville's Theorem (Every Bounded Entire Function is Constant)|Liouville's theorem]] implies that $f$ is in fact constant, which would force $p$ to be constant, which is a contradiction. # Applications **Consequences**: The [[Irreducible Polynomials Over The Complex Numbers|irreducible polynomials over the complex numbers]] are exactly those of degree $1$ and [[Irreducible Polynomials Over The Real Numbers|irreducible polynomials over the real numbers]] are exactly those of degree $1$ or those of degree $2$ of the form $ax^{2}+bx+c$ where $a,b,c\in \mathbb{R}$ with $b^{2}-4ac<0.$