Answer: [tex]\bold{c)\quad \dfrac{2(2x+7)}{(x-4)(x+2)}}[/tex]
Step-by-step explanation:
[tex]\dfrac{x^2+3x+2}{x^2-2x-8}-\dfrac{x^2-x-12}{x^2-2x-8}\\\\\\\text{They already have a common denominator so just add the numerators.}\\\text{Distribute the negative sign to the fraction on the right side and factor}\\\text{the denominator.}\\\\\dfrac{x^2+3x+2-x^2+x+12}{(x-4)(x+2)}=\dfrac{4x+14}{(x-4)(x+2)}=\dfrac{2(2x+7)}{(x-4)(x+2)}[/tex]
