I want to prove that every real polynomial of odd degree has at least one real root, using the intermediate value theorem.
Let $P(x) = x^{2n+1} + a_n x^{2n} + . . . + a_0$ for each $a_i \in \mathbb{R}$ and $n \in \mathbb{N}$.
By the fundamental theorem of algebra I know that $P(x)$ has exactly $2n+1$ complex roots, so
$P(x) = (x+r_1)(x+r_2) . . . (x+r_{2n+1})$ for each $r_i \in \mathbb{C}$
I do not know how to complete this but I do know that, at some point, I probably have to show that each root with imaginary part non zero has to come in conjugate pairs, and since $2n+1$ is odd there is at least $1$ root that is imaginary part $0$ and thus real.
Best Answer
Method of FTA: $$P(\overline z)=\sum_{k=0}^{2n+1}a_k\overline z^k=\sum_{k=0}^{2n+1}\overline a_k\overline{z^k}=\sum_{k=0}^{2n+1}\overline{a_kz^k}=\overline{\sum_{k=0}^{2n+1}a_kz^k}=\overline{P(z)}$$ which states $z$ is a root for $P(z)=0$ iff its complex conjugate $\bar z$ is. According to FTA, there are odd number of roots for a polynomial of odd degree. That implies there must be one single root $z$ satisfying $z=\bar z$, hence the real root.
Method of IVT:
$$\frac{P(x)}{x^{2n+1}}=1+\sum_{k=0}^{2n}a_k\frac{x^k}{x^{2n+1}}=1+\sum_{k=0}^{2n}a_kx^{k-(2n+1)}$$ For any $\varepsilon>0$, there exists $N>0$ such that for all $|x|>N$, $\left|\sum_{k=0}^{2n}a_kx^{k-(2n+1)}\right|<\varepsilon$. Hence for $x>N$, we have $P(x)>x^{2n+1}-\varepsilon x^{2n+1}>0$ and similarly for $x<-N$, we have $P(x)<0$. Then IVT implies there exists some $y$ such that $P(y)=0$.