serie exercices: polynomes, Solution ex 6
📅 February 11, 2024 | 👁️ Views: 218
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& \Centering {Correction Série : Polynomes - Exercice 6} & \hfill TCS \\
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\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}\\
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