Let $P \in \mathbb{K}[X]$ and $(a, b) \in \mathbb{K}^{2}$ such that $a \neq b$. Knowing that the remainder of the Euclidean division of $P$ by $X-a$ is 1 and that the remainder of the Euclidean division of $P$ by $X-b$ is -1, determine the remainder of the Euclidean division of $P$ by $(X-a)(X-b)$.
The translation of the hypotheses allows us to deduce the existence of polynomials $Q_{1}$ and $Q_{2}$ satisfying the following system:

$$
\left\{\begin{array} { l } 
{ P = ( X - a ) Q _ { 1 } + 1 } \\
{ P = ( X - b ) Q _ { 2 } - 1 }
\end{array} \quad \text { hence } \quad \left\{\begin{array}{l}
P(a)=1 \\
P(b)=-1
\end{array}\right.\right.
$$

Let's determine the remainder of the Euclidean division of $P$ by $(X-a)(X-b)$:

$$
P=(X-a)(X-b) Q_{3}+c X+d
$$

where $Q_{3}$ is a certain polynomial (which we do not wish to determine), and $c, d$ are constants to be determined.

Evaluate $(\star)$ at $a$ and $b$ to write the following system:

$$
\left\{\begin{array} { l } 
{ P ( a ) = c a + d } \\
{ P ( b ) = c b + d }
\end{array} \Longleftrightarrow \left\{\begin{array} { r l } 
{ 1 = c a + d } \\
{ - 1 = c b + d } & { L _ { 2 } }
\end{array} \Longleftrightarrow \underset { L _ { 2 } - L _ { 1 } } { } \left\{\begin{array} { r l } 
{ 1 = c a + d } \\
{ - 2 = } & { c ( b - a ) }
\end{array} \underset { a \neq b } { \Longleftrightarrow } \left\{\begin{array}{l}
d=\frac{b+a}{b-a} \\
c=\frac{2}{a-b}
\end{array}\right.\right.\right.\right.
$$

The remainder of the Euclidean division of $P$ by $(X-a)(X-b)$ is given by:

$$
R(X)=c X+d=\frac{2}{a-b} X+\frac{b+a}{b-a}
$$

## Roots

## Link between roots and divisibility

\& Explanation \& We recall that a root $\lambda \in \mathbb{K}$ of $P$ is a value that nullifies $P$, i.e., satisfies $P(\lambda)=0$. It is important to pay attention to the set in which we are looking for the roots, as shown by the following concrete examples.

Example $2 X^{2}-2$ has two real roots $\pm \sqrt{2}$, but has no rational roots (since $\sqrt{2}$ is irrational). $X^{2}+1$ has no roots in $\mathbb{R}$, but has two complex roots $\pm i$.

\& Explanation The following fundamental result allows us to link two distinct worlds (that of evaluation) and that of arithmetic (in terms of divisibility):

## Proposition 1.2.10.

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

$$
\begin{gathered}
\lambda \text { is a root of } P(\text { i.e. } P(\lambda)=0) \Longleftrightarrow P \text { is divisible by }(X-\lambda) \text { (i.e. } \\
\exists Q \in \mathbb{K}[X], P=(X-\lambda) Q)
\end{gathered}
$$

## Proof.

$\Leftrightarrow$ The right-to-left direction is trivial.

$\Rightarrow$ We perform the Euclidean division of $P$ by $(X-\lambda)$. Evaluating at $\lambda$ then allows us to show that its remainder is zero.

## Maximum number of roots

\& Explanation A polynomial cannot have more roots than its degree:

## Theorem 1.2.11.

Let $n \in \mathbb{N}$. A polynomial in $\mathbb{K}[X]$ of degree $n$ has at most $n$ distinct roots in $\mathbb{K}$.

Proof. Induction on $n$ done in class.

Example 3 There exist polynomials that have no real roots, for example:

$$
P(x)=x^{4}+1=\left(x^{2}\right)^{2}+1^{2}+2 x^{2}-2 x^{2}=\left(x^{2}+1\right)^{2}-(\sqrt{2} x)^{2}=\left(x^{2}+\sqrt{2} x+1\right)\left(x^{2}-\sqrt{2} x+1\right)
$$

then we show that the discriminants of these quadratic polynomials are strictly negative (if necessary, we can refer to the section on quadratic polynomial reminders).

Example 4 A polynomial $P \in \mathbb{R}[X]$ of odd degree has at least one real root. It suffices to look at the limits at $\pm \infty$ and conclude by the intermediate value theorem.

## Corollary 1.2.12.

- If a polynomial $P$ has an infinite number of roots, THEN $P$ is zero.
- If a polynomial $P$ of degree $\leqslant n$ has $n+1$ distinct roots, THEN $P$ is zero.
- IF $P=\sum_{k=0}^{n} a_{k} x^{k}$ is of degree $n$ and has $n$ distinct roots $\lambda_{1}, \cdots, \lambda_{n}$, THEN:

$$
P(x)=a_{n}\left(x-\lambda_{1}\right) \cdots\left(x-\lambda_{n}\right)
$$

Proof. The first two points directly follow from the previous theorem. The last point is proven in class by considering the difference.

- Explanation Despite appearances, this corollary, though very intuitive, is one of the most important in this chapter! One of the recipes for solving polynomial exercises consists in:


## INTERPRETING THE HYPOTHESES IN TERMS OF ROOTS OF A CERTAIN POLYNOMIAL