Question

please help
what is the difference in simplest form? (n^2+3n+2/n^2+6n+8)-(2n/n+4)

A. 1/n+2
B. 1-n/n+4
C. 3n+1/n+4
D. n^2+2+2/n^2+6n+8

Answer 1

$\dfrac{n^2+3n+2}{n^2+6n+8}-\dfrac{2n}{n+4}\\\\=\dfrac{n^2+2n+n+2}{n^2+4n+2n+8}-\dfrac{2n}{n+4}\\\\=\dfrac{n(n+2)+1(n+2)}{n(n+4)+2(n+4)}-\dfrac{2n}{n+4}\\\\=\dfrac{(n+2)(n+1)}{(n+2)(n+4)}-\dfrac{2n}{n+4}\\\\=\dfrac{n+1}{n+4}-\dfrac{2n}{n+4}=\dfrac{n+1-2n}{n+4}=\dfrac{1-n}{n+4}$

$Answer:\ \boxed{B.\ \dfrac{1-n}{n+4}}$

Answer 2

Answer:

It is 1-n/n+4.