Solve $2 x^{4}-3 x^{3}-5 x^{2}+7 x-2=0$.

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We first look for integer solutions. If $x$ is such a solution, it divides 2, so $x$ is equal to $\pm 1$ or $\pm 2$. We verify that $x=2$ is indeed a solution. We perform the Euclidean division of $2 x^{4}-3 x^{3}-5 x^{2}+7 x-2$ by $x-2$:

$$
\begin{array}{r|l}
2 x^{4}-3 x^{3}-5 x^{2}+7 x-2 & x-2 \\
\cline { 2 - 2 }-5 x^{2}+7 x-2 & 2 x^{3}+x^{2}-3 x+1 \\
-3 x^{2}+7 x-2 & \\
x-2 & \\
0 &
\end{array}
$$

There is no other integer solution. We then look for rational solutions $p / q$. Since $p$ divides 1 and $q$ divides 2, we try $\pm \frac{1}{2}$ and indeed verify that $\frac{1}{2}$ is a solution. We divide the polynomial by $2 x-1$, which gives $x^{2}+x-1$ whose two roots are $\frac{-1 \pm \sqrt{5}}{2}$. Finally, the roots of the initial polynomial are $2, \frac{1}{2}$ and $\frac{-1 \pm \sqrt{5}}{2}$.

Formally, if $A$ and $B$ are two polynomials with $B \neq 0$, then there exists a unique pair $(Q, R)$ of polynomials such that $A=B Q+R$ with $\operatorname{deg}(R)<\operatorname{deg}(B)$.

$Q$ and $R$ are called the quotient and the remainder of the Euclidean division of $A$ by $B$. Thanks to the Euclidean division, we can define the Euclidean algorithm, the concept of the GCD of two polynomials, establish a Bézout's theorem, etc. We will not go into details.

The Euclidean division of polynomials also allows us to see that

Theorem 120. Let $P$ be a polynomial of degree $n$. Then $P$ has at most $n$ roots.

We reason by induction on the degree. The statement is true for degree 0. Suppose it is true for degree $n-1$. Let $a$ be a root of $P$. We perform the Euclidean division of $P$ by $x-a$:

$$
P(x)=(x-a) Q(x)+b
$$

where $b$ is a real number. Since $P(a)=0$, we have $b=0$ so $P(x)=(x-a) Q(x)$. By the induction hypothesis, $Q$ has at most $n-1$ roots, so $P$ has at most $n$ roots.