Factorize $X^{2}+X+1$ in $\mathbb{C}[X]$ and in $\mathbb{R}[X]$.
$X^{2}+X+1=(X-j)(X-\bar{j})$ with $j=\mathrm{e}^{i \frac{2 \pi}{3}}$ in $\mathbb{C}[X]$ and $X^{2}+X+1=X^{2}+X+1$ in $\mathbb{R}[X]$.

## - Viète's Formulas -

& Explanation & This paragraph continues the idea that knowledge of the roots of $P$ provides information about $P$.

Case $n=2$

Theorem 1.2.17.

Let $P(x)=a x^{2}+b x+c$ be a real polynomial of degree 2 (i.e., $a \neq 0$) with $\lambda_{1}$ and $\lambda_{2}$ as real roots. Then:

$$
\lambda_{1} \lambda_{2}=\frac{c}{a} \quad \text { and } \quad \lambda_{1}+\lambda_{2}=-\frac{b}{a}
$$

Proof. It suffices to expand $P(x)=a\left(x-\lambda_{1}\right)\left(x-\lambda_{2}\right)=a x^{2}+b x+c$ and identify according to the powers of $x$.

## General Case

## Definition 1.2.18.

Let $n \geqslant 1$ and $\lambda_{1}, \cdots, \lambda_{n} \in \mathbb{K}$. For every $k \in \llbracket 1, n \rrbracket$, the $k^{\text {th }}$ elementary symmetric function is the quantity:

$$
\sigma_{k}\left(\lambda_{1}, \cdots, \lambda_{n}\right)=\sum_{1 \leqslant i_{1}<\cdots<i_{k} \leqslant n} \lambda_{i_{1}} \cdots \lambda_{i_{k}}
$$

Example 6 In the case $n=3$, the three elementary symmetric functions of the roots are:

$$
\begin{aligned}
& \sigma_{1}\left(\lambda_{1}, \lambda_{2}, \lambda_{3}\right)=\lambda_{1}+\lambda_{2}+\lambda_{3} \\
& \sigma_{2}\left(\lambda_{1}, \lambda_{2}, \lambda_{3}\right)=\lambda_{1} \lambda_{2}+\lambda_{1} \lambda_{3}+\lambda_{2} \lambda_{3} \\
& \sigma_{3}\left(\lambda_{1}, \lambda_{2}, \lambda_{3}\right)=\lambda_{1} \lambda_{2} \lambda_{3}
\end{aligned}
$$

Theorem 1.2.19.

Let $P(x)=a_{0}+a_{1} x+\cdots+a_{n} x^{n}$ with $a_{n} \neq 0$. Suppose that $P$ has $n$ real roots $\lambda_{1}, \cdots, \lambda_{n}$ (counted with multiplicity). Then, for every $k \in \llbracket 1, n \rrbracket$:

$$
\sigma_{k}\left(\lambda_{1}, \cdots, \lambda_{n}\right)=(-1)^{k} \frac{a_{n-k}}{a_{n}}
$$

In particular:

$$
\sum_{i=1}^{n} \lambda_{i}=-\frac{a_{n-1}}{a_{n}} \quad \sum_{1 \leqslant i<j \leqslant n} \lambda_{i} \lambda_{j}=\frac{a_{n-2}}{a_{n}} \quad \prod_{i=1}^{n} \lambda_{i}=(-1)^{n} \frac{a_{0}}{a_{n}}
$$

Proof. It suffices to expand and identify.