Answer:
[tex]\frac{x+2}{3x-1}[/tex]
Explanation:
→ [tex]\frac{(x^2-9)(x+2)}{(3x^2 +8x -3)(x-3)}[/tex]
→ [tex]\frac{\left(x+3\right)\left(x-3\right)\left(x+2\right)}{\left(3x^2+8x-3\right)\left(x-3\right)}[/tex]
→ [tex]\frac{\left(x+3\right)\left(x+2\right)}{3x^2+8x-3}[/tex]
→ [tex]\frac{\left(x+3\right)\left(x+2\right)}{\left(3x-1\right)\left(x+3\right)}[/tex]
→ [tex]\frac{x+2}{3x-1}[/tex]
