\documentclass[12pt,a4paper]{article}
\usepackage{tabularx}
\usepackage{booktabs}
\usepackage{ragged2e}
\usepackage[left=1.00cm, right=1.00cm, top=0.50cm, bottom=1.00cm]{geometry}
\usepackage{amsmath,amsfonts,amssymb}
\usepackage{fontspec}
\usepackage{mathrsfs}
\usepackage{enumitem}
\usepackage{setspace}
\usepackage{multirow}
\usepackage{xcolor}
\usepackage[ddmmyyyy]{datetime}

\usepackage{tikz,tkz-tab}
\usetikzlibrary{shapes,decorations.text}

\usepackage{polynom}
\usepackage{draftwatermark}
\usepackage{hyperref}

\polyset{style=D}% A,B,C,D
\setstretch{1.5}

\hypersetup{
    colorlinks=true,
    linkcolor=blue
}
\definecolor{cc}{rgb}{236,0,140}
\newcommand{\mylink}{\href{https://mosaid.xyz/cc}{www.mosaid.xyz}}

\SetWatermarkText{\color{cc!10!white}{www.mosaid.xyz}}
\SetWatermarkLightness{0.95}
\SetWatermarkScale{0.8}

\newcolumntype{C}{>{\Centering\arraybackslash}X}
\newcolumntype{A}{>{\centering\arraybackslash}m{1em}}%


\newcommand{\stamp}[2]{
\begin{tikzpicture}[remember picture, overlay]
\coordinate (A) at (#1,#2);
\draw[red!50] (A) circle (1.9cm);
% Draw the inner circle
\draw[red!50] (A) circle (1.4cm);
% Draw the curved line
\draw[red!50, decorate, decoration={text along path,
    text={|\fontspec{DejaVu Sans}\color{red!75}\bfseries|★MOSAID RADOUAN★},
    text align={align=center}, raise=-3pt}] (A) ++ (180:1.6cm) arc (180:0:1.6cm);
\draw[decorate, decoration={text along path,
    text={|\fontspec{DejaVu Sans}\color{red!75}\bfseries|∞★~mosaid.xyz~★∞ },
    text align={align=center}, raise=-6.5pt}] (A) ++ (180:1.53cm) arc (-180:0:1.53cm);
\node[red!75,font=\fontsize{48}{48}\fontspec{DejaVu Sans}\bfseries\selectfont] at (A) {✷};
\end{tikzpicture}
}

\begin{document}
\thispagestyle{empty}
\noindent
\begin{center}
    \begin{tabular}{@{}p{0.22\textwidth}p{0.57\textwidth}p{0.17\textwidth}}
        %\toprule
            \multirow{2}{*}{\parbox{\linewidth}{Prof MOSAID \newline \mylink }}
            & \Centering {Correction Série : Polynomes - Exercice 6} & \hfill  TCS \\
        \bottomrule
    \end{tabular}
\end{center}
\noindent
Soit le polynome \(P(x)=x^3-(a-b)x^2+(a-3b-1)x+2\sqrt2\)\\
{\large\textbullet} 1. Determiner $a$ et $b$  tels que \(P(x)\) est divisible par $x-2$ et $x+\sqrt2$, càd 2 ~et~ $-\sqrt2$ sont des racines \\
On a
$
\begin{cases}
    P(2)=0\\
    P(-\sqrt2)=0
\end{cases}
$
\\éq.à
$
\begin{cases}
    8-(a-b)\times 4 +(a-3b-1)\times 2+2\sqrt2=0\\
    -2\sqrt2-(a-b)\times 2+(a-3b-1)(-\sqrt2)+2\sqrt2=0
\end{cases}
$
\\éq.à
$
\begin{cases}
    a+b-3-\sqrt2=0\\
    (-2-\sqrt2)a+(2+3\sqrt2)b+\sqrt2=0
\end{cases}
$
\\éq.à
$
\begin{cases}
    b=3+\sqrt2-a\\
    (-2-\sqrt2)a+(2+3\sqrt2)\textcolor{red}{(3+\sqrt2-a)}+\sqrt2=0
\end{cases}
$
\\éq.à
$
\begin{cases}
    b=3+\sqrt2-a\\
    (-2-\sqrt2-2-3\sqrt2)a+6+2\sqrt2+9\sqrt2+6+\sqrt2=0
\end{cases}
$
\\éq.à
$
\begin{cases}
    b=3+\sqrt2-a\\
    (-4-4\sqrt2)a+12+12\sqrt2=0
\end{cases}
$
\\éq.à~~~
$a=\dfrac{12+12\sqrt2}{4+4\sqrt2}=3$~~et~~ $b=\sqrt2$\\
{\large\textbullet} 2. On pose $a=3$ ~~;~~$b=\sqrt2$\\
Donc $P(x)=x^3-(3-\sqrt2)x^2+(2-3\sqrt2)x+2\sqrt2$\\
\textbf{2.1} Pour Determiner \(Q(x)\) tel que  \(P(x)=(x-2)Q(x)\),\\
\stamp{15}{4}
On effectue la division euclidienne de \(P(x)\) par \(x-2\)\\
\[
\begin{array}{l|l}
x^3-(3-\sqrt2)x^2+(2-3\sqrt2)x+2\sqrt2 & x-2 \\
\cline{2-2}
x^3 - 2x^2  \hspace*{5cm} & x^2-(1-\sqrt2)x-\sqrt2 \\
\raisebox{1ex}{\rule{3cm}{0.4pt}}  &\\
~~~~(-1-\sqrt2)x^2 + (2-3\sqrt2)x  &\\
~~~~(-1-\sqrt2)x^2 + (2-2\sqrt2)x  &\\
~~~~\raisebox{1ex}{\rule{5cm}{0.4pt}}  &\\
\hspace*{4.3cm}-\sqrt{2}x+2\sqrt2  &\\
\hspace*{4.3cm}-\sqrt{2}x+2\sqrt2  &\\
\hspace*{4.3cm}\raisebox{1ex}{\rule{2.5cm}{0.4pt}} &\\
\hfill 0&\\
\end{array}
\]
Donc $Q(x)=x^2-(1-\sqrt2)x-\sqrt2$\\
\textbf{2.2}. On a $Q(-\sqrt2)=(-\sqrt2)^2-(1-\sqrt2)(-\sqrt2)-\sqrt2=0$ \\
\textbf{2.3}. La division euclidienne de $Q(x)$ par $x+\sqrt2$:\\
\[
\begin{array}{l|l}
x^2-(1-\sqrt2)x-\sqrt2 & x+\sqrt2 \\
\cline{2-2}
x^2 +\sqrt2{x} & x -1\\
\raisebox{1ex}{\rule{3cm}{0.4pt}}  &\\[0.1cm]
\hspace*{2.2cm}-x-\sqrt2  &\\
\hspace*{2.2cm}-x-\sqrt2  &\\
\hspace*{2.2cm}\raisebox{1ex}{\rule{2cm}{0.4pt}}  &\\
\hfill 0~&\\
\end{array}
\]
Alors $Q(x)=(x-1)(x+\sqrt2)$~~ et ~~ $P(x)=(x-2)(x-1)(x+\sqrt2)$\\
\textbf{2.4}. Résoudre $P(x)<0$:\\
\begin{tikzpicture}[scale=1.00]
    \tkzTabInit
    {$x$ /1, $x-2$ /1, $x-1$ /1, $x+\sqrt2$/1,$P(x)$/1}
    {$-\infty$,$-\sqrt2$,$1$,$2$,$+\infty$}
    \tkzTabLine{,-, t, -, t, -, z, +}
    \tkzTabLine{,-, t, -, z, +, t, +}
    \tkzTabLine{,-, z, +, t, +, t, +}
    \tkzTabLine{,-, z, +, z, -, z, +}
\end{tikzpicture}\\
Donc l'ensemble des solutions de l'inéquation est : $S=]-\infty,-\sqrt2[\cup]1;2[$\\
\\
{\large\textbullet} 3. Soit $x\in ]0,1[$. Encadrer $P(x)=x^3-(3-\sqrt2)x^2+(2-3\sqrt2)x+2\sqrt2$ \\
On a $0<x<1$ ~~ donc ~~ $0<x^3<1$ ~~et~~$0<x^2<1$\\
Donc
$
\begin{cases}
    0<x^3<1\\
    -(3-\sqrt2)<-(3-\sqrt2)x^2<0\\
    2-3\sqrt2<(2-3\sqrt2)x<0\\
\end{cases}
$\\[0.5cm]
\stamp{16}{2.3}
Donc~~~~~~~~
$
-1-2\sqrt2~~~~~~~~<x^3-(3-\sqrt2)x^2+(2-3\sqrt2)x~~~~~~~~<1
$\\
Donc~~~~~~~~
$
-1-2\sqrt2\textcolor{red}{+2\sqrt2}<x^3-(3-\sqrt2)x^2+(2-3\sqrt2)x\textcolor{red}{+2\sqrt2}<1 \textcolor{red}{+2\sqrt2}
$\\
Donc~~~~~~~~ $-1<P(x)<1+2\sqrt2$ \\
alors~~~~~~~ $-1+\sqrt2<P(x)-\sqrt2<1+\sqrt2$\\
Donc~~~~~~~~ $|P(x)-\sqrt2|<1+\sqrt2$.~~~~~ la valeur approchée: $\sqrt2$, la précision: $1+\sqrt2$\\
\\
\textcolor{white}{.}\hfill \underline{MOSAID le \today}\\
\textcolor{white}{.}\hfill \mylink
\end{document}