on pose B=4x2-25-(2x+5)(3x-7) 1)Developper et reduire B 2)Apres avoir factoriser 4x2-25, factoriser B 3)Calculer B si x=0, puis x= 2/5

1) B = 4x²-25-(2x+5)(3x-7)
B = 4x²-25-(6x²-14x+15x-35)
B = 4x²-25-6x²+14x-15x+35
B = -2x²-x+10

2) 4x²-25 <=> (2x-5)(2x+5)

B = 4x²-25-(2x+5)(3x-7)
B = (2x-5)(2x+5)-(2x+5)(3x-7)
B = (2x+5)[(2x-5)-(3x-7)]
B = (2x+5)[2x-5-3x+7]
B = (2x+5)(-x+2)

3) Je remplace x par 0 :
B = -2*(0)²-0+10
B = 10

Je remplace x par (2/5)
B = (2*(2/5)+5)(-(2/5)+2)
B = ((4/5)+(25/5))(-(2/5)+(10/5))
B = (29/5)*(8/5)
B = 232/25