Respuesta :

Answer:

see below

Step-by-step explanation:

[tex]\frac{3x^2+7x+4}{x^2+3x+2}\\\\=\frac{(3x+4)(x+1)}{(x+2)(x+1)}\\\\=\frac{3x+4}{x+2}\\\\[/tex]

and the other

[tex]\frac{x^2-2x-8}{3x^2+7x+2}\\\\=\frac{(x-4)(x+2)}{(3x+1)(x+2)}\\\\=\frac{x-4}{3x+1}[/tex]