Factoring the numerator
[tex]n^2-15n+56=0[/tex]use the formula
[tex]n=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]and replace
[tex]\begin{gathered} n=\frac{-(-15)\pm\sqrt[]{(-15)^2-4(1)(56)}}{2(1)} \\ \\ n=\frac{15\pm\sqrt[]{1}}{2} \\ \\ n=\frac{15\pm1}{2} \\ \\ n_1=\frac{15+1}{2}=8 \\ \\ n_2=\frac{15-1}{2}=7 \end{gathered}[/tex]the factored expression is
[tex](n-8)(n-7)[/tex]now replacing on the first expression
[tex]\frac{(n-8)(n-7)}{(n-8)}[/tex]we can simplify (n-8),so
[tex]\frac{(n-8)(n-7)}{(n-8)}\longrightarrow(n-7)[/tex]so, the right option is A
