[tex]=\dfrac{1}{x+2}\cdot\dfrac{x-5}{x-5}+\dfrac{6}{x-5}\cdot\dfrac{x+2}{x+2}\\\\=\dfrac{1(x-5)+6(x+2)}{(x+2)(x-5)}\\\\=\dfrac{7x+7}{x^{2}-3x-10}[/tex]
Usually, the "simplified" form has the polynomial multiplied out (as immediately above), rather than factored. If you want it factored, you can remove a factor of 7 from the terms of the numerator to get 7(x+1)/((x+2)(x-5)).
