serie exercices: polynomes, Solution ex 7
📅 February 12, 2024 | 👁️ Views: 87

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& \Centering {Correction Série : Polynomes - Exercice 7} & \hfill TCS \\
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\noindent
Soit le polynome \(P(x)=x^3-(3\sqrt{3}+1)x^2+m(2+\sqrt{3})x-6\)\\
1. Determiner la valeur de \(m\) tel que \(P(x)\) est divisible par \(x-1\) , càd \(P(1)=0\)\\
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\(1^3-(3\sqrt{3}+1)\times1^2+m(2+\sqrt{3})\times1-6=0\)\\
\(1-(3\sqrt{3}+1)+m(2+\sqrt{3})-6=0\)\\
\((2+\sqrt{3})m=6+3\sqrt{3}\)\\
\(m=\frac{6+3\sqrt{3}}{2+\sqrt{3}}\)\\
\(m=3\)\\
\end{center}
On pose \(m=3\) alors \(P(x)=x^3-(3\sqrt{3}+1)x^2+(6+3\sqrt{3})x-6\)\\
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2.1 Pour Determiner \(Q(x)\) tel que \(P(x)=(x-1)Q(x)\), On effectue la division euclidienne de \(P(x)\) par \(x-1\)\\
\[
\begin{array}{r|l}
x^3 - (3\sqrt{3} + 1)x^2 + (6 + 3\sqrt{3})x - 6 & x-1 \\
\cline{2-2}
\quad x^3 - x^2 \hspace*{5cm} & x^2-3\sqrt{3}x+6 \\
\raisebox{1ex}{\rule{3cm}{0.4pt}} \hspace*{3.5cm} &\\
-3\sqrt{3}x^2 + (6 + 3\sqrt{3})x \hspace*{1cm} &\\
-3\sqrt{3}x^2 + 3\sqrt{3}x \hspace*{2cm} &\\
\raisebox{1ex}{\rule{4cm}{0.4pt}} \hspace*{1cm} &\\
6x-6 &\\
6x-6 &\\
\raisebox{1ex}{\rule{1.2cm}{0.4pt}} &\\
0&\\
\end{array}
\]
Donc \(Q(x)=x^2-3\sqrt{3}x+6\)\\
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3. Pour vérifier si \(\sqrt{3}\) est une racine du polynôme \(Q(x)\), nous devons substituer \(x = \sqrt{3}\) dans \(Q(x)\) et voir si le résultat est égal à zéro.
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\begin{center}
\begin{align*}
Q(\sqrt{3}) &= (\sqrt{3})^2 - 3\sqrt{3}\cdot\sqrt{3} + 6 \\
&= 3 -3\times3+ 6 \\
&= 0
\end{align*}
\end{center}
Comme le résultat est \(0\), cela signifie que \(\sqrt{3}\) est une racine du polynôme \(Q(x)\).\\
4. {\large\textbullet}Diviser \(Q(x)\) par \(x-\sqrt{3}\)\\
\vspace*{-0.7cm}
\[
\begin{array}{r|l}
x^2 - 3\sqrt{3}x + 6 & x-\sqrt{3} \\
\cline{2-2}
\quad x^2 - \sqrt{3}x \hspace*{0.9cm} & x-2\sqrt{3} \\
\raisebox{1ex}{\rule{2cm}{0.4pt}} \hspace*{0.8cm} &\\
-2\sqrt{3}x +6&\\
-2\sqrt{3}x +6&\\
\raisebox{1ex}{\rule{2.2cm}{0.4pt}} &\\
0&\\
\end{array}
\]
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5. Factoriser \(P(x)\):
\(P(x)=(x-1)(x-\sqrt{3})(x-2\sqrt{3})\)\\
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