Bonjour,
[tex]A(x)=(2x+1)(x-5)\\
A(x)=2x^2-10x+x-5\\
\boxed{A(x)=2x^2-9x-5}\\\\
A(-3)=2\times(-3)^2-9\times(-3)-5\\
A(-3)=18+27-5\\
\boxed{A(-3)=40}\\\\
A(x) = 0\\
(2x+1)(x-5)=0\\
2x+1=0 \ \ ou \ \ x-5=0\\
\boxed{x= \frac{-1}{2} \ \ ou \ \ x=5}[/tex]
