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\noindent
\begin{center}
    \begin{tabular}{@{}p{0.22\textwidth}p{0.57\textwidth}p{0.17\textwidth}}
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            \multirow{2}{*}{\parbox{\linewidth}{Prof MOSAID \newline \mylink }}
            & \Centering {Correction Série : Polynomes - Exercice 5} & \hfill  TCS \\
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\vspace*{0.4cm}
Soit $P(x)=x^3-2x^2-5x+6$\\
{\large\textbullet} 1. Pour montrer que $P(x)$ est divisible par $x-1$. il suffit de vérifier que $P(1)=0$\\
{\large\textbullet} 2. Effectuer la division euclidienne de $P(x)$ par $x-1$:\\
\stamp{3}{-3}
\[ \polylongdiv{x^3-2x^2-5x+6}{x-1} \] \\
Donc $P(x)=(x-1)(x^2-x-6)$~~ et~~ $Q(x)=x^2-x-6$\\
{\large\textbullet} 3. Pour factoriser $Q(x)$ on peut utiliser le discriminant pour trouver les deux racines de $Q(x)$ (dans un cours prochain). On peut aussi remarquer que :
\begin{align*}
    x^2-x-6&=x^2-3x+2x-6\\
           &=x(x-3)+2(x-3)\\
           &=(x-3)(x+2)
\end{align*}
Donc $P(x)=(x-1)(x-3)(x+2)$
\\
{\large\textbullet} 4. Pour résoudre l'inéquation $P(x)<0$ on dress le tableau de signe du produit:\\
\begin{tikzpicture}[scale=1.00]
    \tkzTabInit
    {$x$ /1, $x-1$ /1, $x-3$ /1, $x+2$/1,$P(x)$/1}
    {$-\infty$,$-2$,$1$,$3$,$+\infty$}
    % Signs for (x - 1)
    \tkzTabLine{,-, t, -, z, +, t, +}
    % Signs for (x - 3)
    \tkzTabLine{,-, t, -, t, -, z, +}
    % Signs for (x + 2)
    \tkzTabLine{,-, z, +, t, +, t, +}
    % Final sign for P(x)
    \tkzTabLine{,-, z, +, z, -, z, +}
\end{tikzpicture}\\
Donc l'ensemble des solutions de l'inéquation est : $S=]-\infty,-2[\cup]1;3[$\\

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