Soit l'expression A(x)=4x[auCarré]-1(2x+1)(x+3) 1. Calculer A(-1/2) et A(0) DM pour demain! Merci a ceux qui repondrons

Salut,

[tex]A(x) = 4 x^{2} -1(2x+1)(x+3)\\ A( \frac{-1}{2} )= 4*\frac{-1}{2}^{2} -1(2*\frac{-1}{2} + 1 )(\frac{-1}{2}+3) \\ A(\frac{-1}{2}) = 4* \frac{1}{4} - 1 (-1+1)(\frac{-1}{2}+3)\\ A(\frac{-1}{2}) = 1 - 1*0 \\ A(\frac{-1}{2}) = 1[/tex]
[tex]A(0)= 4*0^{2} -1(2*0+1)(0+3) \\ A(0) = 0 - 1*1*3 \\ A(0) = -3[/tex]
Bonne soirée !

A(x)=4x[auCarré]-1(2x+1)(x+3) =
A(x) = ax² - 1( 2x+1) ( x+3)

A( -1/2)
A( -1/2) = 4* (-1/2)² - 1 (2 * (-1/2) +1) ( -1/2 +3)
A (-1/2) = 4*1/4 - 1 (-1+1) ( -1/2+3)
A(-1/2) =  1

A(0)
A(0) = 4* 0² - 1 (2*0 +1) (0+3)
A (0) = 0 - 1*1*3
A(0) = -3