Développements, factorisation (Prérentrée)

Exercice 1 (🔥) : Développements de produits
📄 Énoncé

Développer et ordonner selon les puissances décroissantes :

  1. \(\left(2 x-\frac{1}{2}\right)^3\)

  2. \((x+1)^2(x-1)\)

  3. \((x+1)^2 \left(x^2-x+1\right)\)

  4. \((2 x+3)(5 x-8)-(2 x-4)(5 x-1)\)

  5. \(\left(x^2+x+1\right)^2\)

  6. \((x^2-2)(2x+3)^2\)

✅ Corrigé

  1. \(\displaystyle \left(2 x-\frac{1}{2}\right)^3= 8x^3 -6x^2+\frac{3}{2}\, x - \frac{1}{8}\)

  2. \((x+1)^2(x-1) = x^3 + x^2-x-1\)

  3. \((x+1)^2 \left(x^2-x+1\right)= x^4+x^3+x+1\)

  4. \((2 x+3)(5 x-8)-(2 x-4)(5 x-1) = 21x-28\)

  5. \(\left(x^2+x+1\right)^2= x^4 + 2x^3+3x^2+2x+1\)

  6. \((x^2-2)(2x+3)^2 = 4x^4+12x^3 + x^2-24x-18\)

Détail des calculs

  1. On a :

    Commentaire

    \((a+b)^3=a^3+3a^2b+3ab^2+b^3\)

    \[\begin{aligned} \left(2 x-\frac{1}{2}\right)^3 &= (2x)^3 - \frac{3}{2} \left( 2x \right)^2 + \frac{3}{4} \left( 2x \right) - \frac{1}{8} \\ &= 8x^3 -6x^2+\frac{3}{2}\, x - \frac{1}{8} \end{aligned}\]

  2. On a :

    Commentaire

    \((a+b)^2=a^2 + 2ab + b^2\)

    \[\begin{aligned} (x+1)^2(x-1) &= (x^2+2x+1)(x-1) \\ &= x^3+2x^2+x-x^2-2x-1 \\ &= x^3 + x^2-x-1 \end{aligned}\]

  3. On a : \[\begin{aligned} (x+1)^2 \left(x^2-x+1\right) &= (x^2+2x+1) \left(x^2-x+1\right) \\ &= x^4-x^3+x^2+2x^3-2x^2+2x+x^2-x+1 \\ &= x^4+x^3+x+1 \end{aligned}\]

  4. On a : \[\begin{aligned} (2 x+3)(5 x-8)-(2 x-4)(5 x-1) &= (10x^2-16x+15x-24)-(10x^2-2x-20x+4) \\ &= 21x-28 \end{aligned}\]

  5. On a :

    Commentaire

    \((a+b+c)^2 = a^2+b^2+c^2+2ab+2ac+2bc\)

    \[\begin{aligned} \left(x^2+x+1\right)^2 &= x^4 + x^2 + 1 + 2x^3+2x^2+2x \\ &= x^4 + 2x^3+3x^2+2x+1 \end{aligned}\]

  6. On a : \[\begin{aligned} (x^2-2)(2x+3)^2 &= (x^2-2) (4x^2+12x+9) \\ &= 4x^4+12x^3+9x^2-8x^2-24x-18 \\ &= 4x^4+12x^3 + x^2-24x-18 \end{aligned}\]

📝 Mes notes :

Exercice 2 (🔥) : Factorisation
📄 Énoncé

Factoriser le plus possible les expressions suivantes :

  1. \(9 x^2-49-(3 x+7)(6 x-1)\)

  2. \((-5 x+3)^2 - 36\).

  3. \((3 x-4)(2 x-5)+9 x^2-16\)

  4. \((8-12x)( x+1)+36 x^2-16\)

  5. \(2x^2-18+(2x-1)(3-x)\)

  6. \(x^2-4xy+4y^2+(x-y)(x-2y)\)

✅ Corrigé

  1. \(9 x^2-49-(3 x+7)(6 x-1) = -3 \left( 3x+7 \right) (x+2)\)

  2. \((-5 x+3)^2 - 36 = (5x+3)(5x-9)\).

  3. \((3 x-4)(2 x-5)+9 x^2-16= (3x-4)(5x-1)\)

  4. \((8-12x)( x+1)+36 x^2-16 = 4 \left( 3x-2 \right) (2x+1)\)

  5. \(2x^2-18+(2x-1)(3-x) = 7 \left( x-3 \right)\)

  6. \(x^2-4xy+4y^2+(x-y)(x-2y) = (x-2y)(2x-3y)\)

Détail des calculs

  1. On a :

    Commentaire

    \(a^2-b^2 = (a-b)(a+b)\)

    \[\begin{aligned} 9 x^2-49-(3 x+7)(6 x-1) &= (3x)^2 - 7^2 - (3 x+7)(6 x-1) \\ &= (3x-7)(3x+7) - (3 x+7)(6 x-1) \\ &= (3x+7)((3x-7) - (6x-1)) \\ &= (3x+7)(-3x-6) \\ &= -3 \left( 3x+7 \right) (x+2) \end{aligned}\]

  2. On a : \[\begin{aligned} (-5x+3)^2 - 36 &= (-5x+3)^2- 6^2 \\ &= (-5x+3)^2- 6^2 \\ &= (-5x+3-6)(-5x+3+6) \\ &=(-5x-3)(-5x+9) \\ &= (5x+3)(5x-9) \end{aligned}\]

  3. On a : \[\begin{aligned} (3 x-4)(2 x-5)+9 x^2-16 &= (3 x-4)(2 x-5)+ (3x)^2 - 4^2 \\ &= (3 x-4)(2 x-5)+ (3x-4)(3x+4) \\ &= (3x-4)(2x-5+3x+4) \\ &= (3x-4)(5x-1) \end{aligned}\]

  4. On a : \[\begin{aligned} (8-12x)( x+1)+36 x^2-16 &= 4 \left[ \left( 2-3x \right)(x+1) + 9x^2-4 \right] \\ &= 4 \left[ (2-3x)(x+1) + (3x-2)(3x+2) \right] \\ &= 4 \left( 3x-2 \right) \left[ - (x+1) + (3x+2) \right] \\ &= 4 \left( 3x-2 \right) (2x+1) \end{aligned}\]

  5. On a : \[\begin{aligned} 2x^2-18+(2x-1)(3-x) &= 2 \left( x^2-9 \right) + (2x-1)(3-x) \\ &= 2 \left( x-3 \right)(x+3) -(2x-1)(x-3) \\ &= (x-3) \left( 2x+6-2x+1 \right) \\ &= 7 \left( x-3 \right) \end{aligned}\]

  6. On a : \[\begin{aligned} x^2-4xy+4y^2+(x-y)(x+2y) &= (x-2y)^2 + (x-y)(x-2y) \\ &= (x-2y) \left[ (x-2y) + (x-y) \right] \\ &= (x-2y)(2x-3y) \end{aligned}\]

📝 Mes notes :

Exercice 3 (🔥🔥) : Factorisation
📄 Énoncé

Factoriser le plus possible les expressions suivantes :

  1. \((x+2)^2-y^2\)

  2. \(1+x -y-x y\)

  3. \(4x^2+4 x y+ y^2- 49 x^2\)

  4. \(x^3+6x^2 y+9xy^2-x^3y^2\)

  5. \(x+y+2\sqrt{xy} - 1\)

  6. \(x^2\left(y^2+z^2\right) - 9x^4y^2-9x^4z^2\)

✅ Corrigé

  1. \((x+2)^2-y^2 = (x-y+2)(x+y+2)\)

  2. \(1+x -y-x y = (1+x)(1-y)\)

  3. \(4x^2+4 x y+ y^2- 49 x^2= (9x+y)(y-5x)\)

  4. \(x^3+6x^2 y+9xy^2-x^3y^2= x \left( x+3y-xy \right) (x+3y+xy)\)

  5. \(x+y+2\sqrt{xy} - 1= (\sqrt{x} + \sqrt{y}-1) (\sqrt{x} + \sqrt{y}+1)\)

  6. \(x^2\left(y^2+z^2\right) - 9x^4y^2-9x^4z^2= x^2 \left( 1-3x \right)(1+3x) (y^2+z^2 )\)

Solutions détaillées

Détail des calculs

  1. On a : \[\begin{aligned} (x+2)^2-y^2 &= (x+2-y)(x+2+y) \end{aligned}\]

  2. On a : \[\begin{aligned} 1+x -y-x y &= 1+x - y \left( 1+x \right) \\ &= (1+x)(1-y) \end{aligned}\]

  3. On a : \[\begin{aligned} 4x^2+4 x y+ y^2- 49 x^2 &= (2x+y)^2 - (7x)^2 \\ &= (2x+y+7x)(2x+y-7x) \\ &= (9x+y)(y-5x) \end{aligned}\]

  4. On a: \[\begin{aligned} x^3+6x^2 y+9xy^2-x^3y^2 &= x \left( x^2+6xy+9y^2 -( xy)^2 \right) \\ &= x \left[ ( x+3y)^2 - (x y)^2 \right] \\ &= x \left( x+3y-xy \right) (x+3y+xy) \end{aligned}\]

  5. On a : \[\begin{aligned} x+y+2\sqrt{xy} - 1 &= (\sqrt{x})^2 + 2 \sqrt{x} \sqrt{y} + (\sqrt{y})^2-1 \\ &= (\sqrt{x} + \sqrt{y})^2 -1 \\ &= (\sqrt{x} + \sqrt{y}-1) (\sqrt{x} + \sqrt{y}+1) \end{aligned}\]

  6. On a : \[\begin{aligned} x^2\left(y^2+z^2\right) - 9x^4y^2-9x^4z^2 &= x^2\left(y^2+z^2- 9x^2y^2-9x^2z^2\right) \\ &= x^2\left[ y^2 \left( 1 - 9x^2 \right) + z^2 \left( 1 - 9x^2 \right) \right] \\ &= x^2 \left( 1-9x^2 \right) (y^2+z^2 ) \\ &= x^2 \left( 1-3x \right)(1+3x) (y^2+z^2 ) \end{aligned}\]

📝 Mes notes :

Chapitre Précédent
Retour à la/au Leçon
Chapitre Suivant