Question

What is the difference in simplest form?(n^2+10n+21) / (n^2+3n-28) - (3n)/(n-4)a) (2n+3)/(n-4)b) n/(n-4)c) (-2n+3)/(n-4)d) (4n+3)/(n-4)

Answer 1

Given the expression:

$\frac{n^2+10n+21}{n^2+3n-28}-\frac{3n}{n-4}$

We can express the quadratic polynomials as:

$\begin{gathered} n^2+10n+21=(n+3)(n+7) \\ n^2+3n-28=(n+7)(n-4) \end{gathered}$

Then:

$\begin{gathered} \frac{(n+3)(n+7)}{(n+7)(n-4)}-\frac{3n}{n-4}=\frac{n+3}{n-4}-\frac{3n}{n-4} \\ \Rightarrow=\frac{n+3-3n}{n-4}=\frac{-2n+3}{n-4} \end{gathered}$