Find the remainder of the Euclidean division of $x^{100}-2 x^{51}+1$ by $x^{2}-1$.

## - Roots and factorization of polynomials

Here we see that the knowledge of the roots of a polynomial allows us to factorize it. Recall that $\mathbb{K}$ denotes $\mathbb{R}$ or $\mathbb{Q}$.

Definition 1.10. An element $x \in \mathbb{K}$ is called a root of a polynomial $P \in \mathbb{K}[X]$ if $P(x)=0$.

Example 1.11. The real polynomial $X^{2}-1$ has two real roots, which are 1 and -1. The polynomial $X^{2}+1$ has no real roots. The real polynomial $X^{2}-2$ has two real roots, but the polynomial with rational coefficients $X^{2}-2$ has no rational roots because $\sqrt{2}$ is irrational.

Theorem 1.12. Let $P \in \mathbb{K}[X]$ and $a \in \mathbb{K}$. The following two propositions are equivalent:

1. a is a root of $P$, that is, $P(a)=0$.
2. There exists a polynomial $Q \in \mathbb{K}[X]$ such that:

$$
P(x)=Q(x)(x-a) \text {. }
$$

Proof. It is clear that the second point implies the first. As for the converse, the key is to use Euclidean division. Indeed, suppose that $P(a)=0$. Then write the Euclidean division of $P$ by $X-a$ in the form $P(x)=Q(x)(x-a)+R(x)$ with $R$ a polynomial of degree at most $1-1=0$. Thus, $R$ is a real number, denoted by $c$. In short, $P(x)=Q(x)(x-a)+c$. Evaluate this quantity at $x=a: 0=P(a)=Q(a)(a-a)+c$. Therefore, $c=0$, which is what we wanted to prove.

Theorem 1.13. A polynomial of degree $n$ has at most $n$ different roots.

Proof. By contradiction, let $P \in \mathbb{R}[X]$ of degree $n$ having at least $n+1$ different roots, denoted $r_{1}, \ldots, r_{n+1}$. According to the previous theorem, there exists a polynomial $Q$ such that $P(x)=$ $Q(x)\left(x-r_{1}\right)$. But then, for $2 \leq i \leq n+1,0=P\left(r_{i}\right)=Q\left(r_{i}\right)\left(r_{i}-r_{1}\right)$. Since $r_{i}-r_{1} \neq 0$, this implies $Q\left(r_{i}\right)=0$. We repeat this reasoning with $Q$ to finally obtain the existence of a polynomial $T$ such that $P(x)=T(x)\left(x-r_{1}\right) \cdots\left(x-r_{n+1}\right)$. Then, according to Proposition 1.6:

$$
n=\operatorname{deg}(P)=\operatorname{deg} T+n+1>n
$$

which is absurd.

Remark 1.14. There exist polynomials that have no real roots, for example $P(x)=$ $x^{4}+1$.

This important theorem implies some corollaries giving information about the polynomial given something about its roots.

Corollary 1.15. Let $P(x)=a_{0}+a_{1} x+\cdots+a_{n} x^{n}$ be a polynomial of degree $n$. Suppose it has $n$ different roots $r_{1}, \ldots, r_{n}$. Then:

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

Proof. By revisiting the previous proof, we see that there exists a polynomial $T$ such that $P(x)=T(x)\left(x-r_{1}\right) \cdots\left(x-r_{n}\right)$. By comparing the degrees of the terms on the left and right, it follows that $T$ is of degree zero, hence a real number. By looking at the leading coefficient on both sides of the equality, we find that $T(x)=a_{n}$.

Corollary 1.16. A polynomial of degree $n$ having $n+1$ roots is zero. Thus, a polynomial having an infinite number of roots is necessarily the zero polynomial.
We are looking for the remainder in the form $R(X)=a X+b$. We have $R(1)=P(1)$, $R(-1)=P(-1)$, which allows us to calculate $R(X)=-2 X+2$.