Let $P \in \mathbb{R}[X]$. Suppose that $P$ is of degree $n$ such that for all $k \in \llbracket 1, n+1 \rrbracket$, $P(k)=1 / k$. Calculate $P(-1)$.
Consider the polynomial $x P(x)-1$, which is of degree $n+1$ and has $n+1$ distinct roots, which are the integers from 1 to $n+1$. We deduce that there exists a constant $K \in \mathbb{K}$ such that, for all $x \in \mathbb{R}$:

$$
P(x)=K \prod_{k=1}^{n+1}(x-k)
$$

Evaluating at 0 allows us to find that $K=\frac{(-1)^{n}}{(n+1)!}$, and then we deduce the expression of $x P(x)$:

$$
x P(x)=1+\frac{(-1)^{n}}{(n+1)!} \prod_{k=1}^{n+1}(x-k)
$$

Evaluating the previously obtained expression at -1 allows us to find that $P(-1)=$ $n+1$.

## Multiple Roots

## Definition 1.2.13.

Let $P \in \mathbb{K}[X]$ be NON ZERO, $\lambda$ a SCALAR, and $m \in \mathbb{N}$.

$$
\begin{aligned}
\lambda \text { is a ROOT OF MULTIPLICITY } m \text { IN } P & \underset{\text { def. } 1}{ }\left[\begin{array}{c}
\exists Q \in \mathbb{K}[X], P=(X-\lambda)^{m} Q \\
\text { AND } \\
(X-\lambda)^{m+1} \text { does not divide } P
\end{array}\right] \\
& \underset{\text { def. } 2}{ }\left[\begin{array}{c}
\exists Q \in \mathbb{K}[X], P=(X-\lambda)^{m} Q \\
\text { AND } \\
Q(\lambda) \neq 0
\end{array}\right]
\end{aligned}
$$

In particular, we have the following characterizations:

$$
\begin{aligned}
\lambda \text { is a root of multiplicity } 0 & \Longleftrightarrow \lambda \text { is not a root of } P \\
\lambda \text { is a root of multiplicity } 1 & \Longleftrightarrow \lambda \text { is a simple root of } P \\
\lambda \text { is a root of multiplicity } 2 & \Longleftrightarrow \lambda \text { is a double root of } P \\
\lambda \text { is a root of multiplicity } m \geqslant 2 & \Longleftrightarrow \lambda \text { is a multiple root of } P
\end{aligned}
$$

Example $5 P=(X-1)^{2}(X+1)(X-\pi)^{17}$ has roots the real numbers $1, -1$, and $\pi$ with respective multiplicities 2, 1, and 17.

## Theorem 1.2.14.

Let $P \in \mathbb{K}[X]$ and $\alpha \in \mathbb{K}$. Then:

$$
\alpha \text { is a multiple root of } P \text { if and only if }\left(P(\alpha)=0 \text { and } P^{\prime}(\alpha)=0\right) \text {. }
$$

Proof. Proof done in class.

& Explanation & More generally, it can be shown by induction that:

Theorem 1.2.15.

Let $P \in \mathbb{K}[X], \alpha \in \mathbb{K}$, and $n$ an integer greater than or equal to 1. Then:

$$
P(\alpha)=P^{\prime}(\alpha)=P^{\prime \prime}(\alpha)=\cdots=P^{(n)}(\alpha)=0 \Longleftrightarrow(X-\alpha)^{n} \text { divides } P
$$

Proof. Left to the reader.