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Polynôme Bac L 2011

Petit Koldois 2 Min Read

Soit le polynôme $P(x) = x^3 + ax^2 + bx + 6$ où a et b sont des réels.

1) Déterminer les réels a et b sachant que P(-2) = 0 et P (-1) = 8.

2) On pose $P(x)= x^3 - 2x^2 - 5x + 6$ .

a) Factoriser P(x).

b) Résoudre dans $\mathbb{R}$ l’équation P(x) = 0.

c) Résoudre dans $\mathbb{R}$ l’inéquation $P(x) \geq 0$ .

d) Déduire de la question 2.b) les solutions de l’équation (E).

E) $e^{3x+1} - 2 e^{2x+1} - 5 e^{x+1} + 6 e = 0$ .

Corrigé

$P(x)=x^3+ax^2+bx+6$ $(a,b) \in \mathbb{R}^2$

1. Déterminons les réels a et b sachant que P(-2)=0 et P(-1)=8

P(-2) = 0 signifie que $(-2)^3+a(-2)^2+b(-2)+6=0$

ce qui donne -8 + 4a – 2b + 6 = 0

soit 4a – 2b – 2 = 0 (1)

P(-1) = 0 signifie que $(-1)^3+a(-1)^2+b(-1)+6=0$

ce qui donne -1 + a – b + 6 = 8

soit a – b = 3 (2)

P(-2) = 0 et P(-1) = 8 se traduit donc par le système :

$\left\{ \begin{array}{l} 4a - 2b = 2 \\ a - b = 3 \end{array} \right.$

soit
$\left\{ \begin{array}{l} 2a - b = 1 \\ a - b = 3 \end{array} \right.$

soit
$\left\{ \begin{array}{l} 2a - b = 1\ (1) \\ a - b = 3\ (2) \end{array} \right.$

soit
$\left\{ \begin{array}{l} a = -2\ (1)-(2) \\ -2 - b = 3 \end{array} \right.$

soit
$\left\{ \begin{array}{l} a = -2\\ b = -5 \end{array} \right.$

2)

a) Factorisons $P(x)=x^3+ax^2+bx+6$ .

Comme P(-2) = 0, alors -2 est racine de P(x) donc $P(x)=(x+2)(x^2+\alpha x +\beta)$

$P(x)= x^3 +\alpha x^2 + 2 x^2 + \beta x + 2\alpha x + 2 \beta$

$P(x)= x^3 +(\alpha + 2)x^2 + (\beta + 2\alpha)x + 2 \beta$

mais $P(x)=x^3+ax^2+bx+6$ donc par identification on a :

$\left\{ \begin{array}{l} \alpha + 2 = -2\\ 2\alpha + \beta = -5 \\ 2\beta = 6 \end{array} \right.$

soit
$\left\{ \begin{array}{l} \alpha = -4\\ \beta = 3 \end{array} \right.$

Ainsi $P(x)=(x+2)(x^2 -4x + 3)$

1 est racine évidente de $x^2 -4x + 3$

donc P(x) = (x+2)(x-2)(x-3)

b) Résolvons l’équation P(x) = 0

$P(x)=0\ \Longleftrightarrow (x+2)(x-2)(x-3) = 0$

$\Longleftrightarrow x = -2\ ou\ x = 2\ ou\ x = 3$

$S=\{-2,2,3\}$

c) Déduisons de 2b) les solutions de (E) :

$(E)\ :\ e^{3x+1} -2e^{2x+1} -5e^{x+1} +6e = 0$

$(E)\Longleftrightarrow e \left(e^{3x} -2e^{2x} -5e^{x} + 6\right) = 0$

en posant $e^{x}=X$ on obtient avec X > 0

$(E)\Longleftrightarrow X^{3} -2X^{2} -5X + 6 = 0$

$\Longleftrightarrow P(X)=0$

$\Longleftrightarrow X = 2\ ou\ X = 3$ car X > 0

Soit $e^{x}=2\ ou\ e^{x}=3$

Soit $x=\ln 2\ ou\ x =\ln 3$

$S' =\{\ln 2, \ln 3\}$

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Petit Koldois 24 mai 2022

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