> [!NOTE] Lemma > Every *odd degree [[Polynomial|polynomial]]* $P(x) \in \mathbb{R}[x]_{2n+1}$, given by $P(x) = \sum_{j=0}^{2n+1} a_{j} x^{j}$ where $a_{2n+1} \neq 0$, has at least one real root. ###### Proof using IVT WLOG, take $a_{2n+1}>0$, since $P(x)=0 \iff -P(x)=0$. Define $A = \sum_{j=0}^{2n} |a_{j}|$. Then for $x>1$ we have $P(x) > a_{2n+1} x^{2n+1} - \left( \sum_{j=0}^{2n}|a_{j}| \right) x^{2n} = x^{2n}(a_{2n+1}x-A).$which is positive if $x> \max\left( 1, \frac{A}{a_{2n+1}} \right)$. Similarly for $x<-1$ we have $P(x) = -P(-x) < -x^{2n}(a_{2n+1}x-A)$which is negative if $x<\min\left( -1,-\frac{A}{a_{2n+1}} \right)$. So there exists $x_{1}>0$ with $P(x_{1})>0$ and $x_{2}<0$ with $P(x_{2})<0$. We have already seen that $P$ is a [[Continuous Real Function|cts]] on $\mathbb{R}$ [[Algebra of Continuous Real Functions|here]], so applying [[Intermediate Value Theorem (IVT)|IVT]] to $f:[x_{2},x_{1}] \to \mathbb{R}$ we find $c \in (x_{2},x_{1})$ such that $P(c)=0$. ###### Proof using IVT [^1] $ \frac{P(x)}{x^{2 n+1}}=1+\sum_{k=0}^{2 n} a_k \frac{x^k}{x^{2 n+1}}=1+\sum_{k=0}^{2 n} a_k x^{k-(2 n+1)} $ Since $0 = \lim_{ x \to \infty } \sum_{k=0}^{2 n} a_k x^{k-(2 n+1)}$ (see [[Limit at Infinity of Univariate Real Function|limit at infinity]]), for any $\varepsilon>0$, there exists $N>0$ such that for all $|x|>N,\left|\sum_{k=0}^{2 n} a_k x^{k-(2 n+1)}\right|<\varepsilon$. Hence for $x>N$, choosing $\varepsilon< 1$ yields $P(x)>x^{2 n+1}(1-\varepsilon)>0$ and similarly for $x<-N$, we have $P(x) < x^{2n+1} (1+\varepsilon) <0$. Then IVT implies there exists some $y$ such that $P(y)=0$. ###### Proof using FTA [^1] Note that $P(z)=0$ iff $P(\bar{z})=0$. According to [[Fundamental Theorem of Algebra|FTA]] there are $2n+1$ complex roots of $P$. Since $2n+1$ is odd there must exist at least one root satisfying $z = \bar{z}$. # References [^1]: https://math.stackexchange.com/questions/689575/proof-that-every-polynomial-of-odd-degree-has-one-real-root