Exercice 3 On considère l'expression A(x) = (x + 1)(x + 3) - (x + 1)(2x + 5) 1. Développer et réduire A(x). 2. Factoriser A(x). 3. Montrer que lorsqu'on remplac

Réponse :

1. A(x) = -x²-3x-2

2. A(x) = -(x+1)(x+2)

3. A(-2) = 0

Explications étape par étape :

1. A(x) = (x+1)(x+3)-(x+1)(2x+5)

  A(x) = x*x+x*3+1*x+1*3-(x*2x+x*5+1*2x+1*5)

  A(x) = x²+3x+x+3-(2x²+5x+2x+5)

  A(x) = x²+(3+1)x+3-(2x²+(5+2)x+5)

  A(x) = x²+4x+3-(2x²+7x+5)

  A(x) = x²+4x+3-2x²-7x-5

  A(x) = x²-2x²+4x-7x+3-5

  A(x) = (1-2)x²+(4-7)x+(3-5)

  A(x) = -1x²+(-3)x+(-2)

  A(x) = -x²-3x-2

2. A(x) = (x+1)(x+3)-(x+1)(2x+5)

   A(x) = (x+1)[(x+3)-(2x+5)]

   A(x) = (x+1)(x+3-2x-5)

   A(x) = (x+1)(x-2x+3-5)

   A(x) = (x+1)((1-2)x+(3-5))

   A(x) = (x+1)((-1)x+(-2))

   A(x) = (x+1)(-1x-2)

   A(x) = (x+1)(-1*x+(-1)*2)

   A(x) = (x+1)(-1)(x+2)

   A(x) = (-1)(x+1)(x+2)

   A(x) = -(x+1)(x+2)

3. Utilisons la forme factorisée de A(x) lorsque x=-2

   A(x) = -(x+1)(x+2)

   A(-2) = -(-2+1)(-2+2)

   A(-2) = -(-1)*(0)

   A(-2) = 1*0

   A(-2) = 0