[tex]\bf \cfrac{n^2+3n+21}{n^2+6n+8}-\cfrac{2n}{n+4}\implies \cfrac{n^2+3n+21}{(n+4)(n+2)}-\cfrac{2n}{n+4}
\\\\\\
\textit{so our LCD will just be (n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+21~~~~-~~~~(n+2)(2n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+21~~~~-~~~~(2n^2+4n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+21~~~~-~~~~2n^2-4n}{(n+4)(n+2)}
\implies
\cfrac{-n^2-n+21}{(n+4)(n+2)}[/tex]
