Hello,
We suppose x≠-2.
[tex] \dfrac{x^2+5x-2}{x+2}=3\\\\
==\textgreater \ \dfrac{x^2+5x-2-3(x+2)}{x+2}=0\\\\
==\textgreater \ x^2+4x-2x-8=0\\
==\textgreater \ x^2+4x-2x-8=0\\
==\textgreater \ x(x+4)-2(x+4)=0\\
==\textgreater \ (x+4)(x-2)=0\\
==\textgreater \ x=-4 \ or \ x=2\\\\
Proof:\\
if\ x=-4\ then \ \ \dfrac{(-4)^2+5*(-4)-2}{-4+2} =\frac{16-20-2}{-2}=\frac{-6}{-2}=3\\\\
if\ x=2\ then \ \ \dfrac{(2)^2+5*(2)-2 }{2+2} =\frac{4+10-2}{4}=\frac{12}{4}=3\\\\
[/tex]
