Question

Simplify the expression $\frac{4}{n^2+6n+9}-\frac{1}{n^2-9}$and explain all your steps.

Short answer

The expression $\frac{4}{n^2+6n+9}-\frac{1}{n^2-9}$is simplified as $\frac{3n-15}{(n+3)(n+3)(n-3)}$.

Step-by-step solution

Step 1. Given Information

We are given an expression $\frac{4}{n^2+6n+9}-\frac{1}{n^2-9}$.

We need to simplify the expression.

Step 2. Find the LCD

In the expression $\frac{4}{n^2+6n+9}-\frac{1}{n^2-9}$, the denominators can be factored as

role="math" localid="1653585283061" $\frac{4}{n^2+6n+9}-\frac{1}{n^2-9}=\frac{4}{(n+3)(n+3)}-\frac{1}{(n+3)(n-3)}$

So the LCD is $(n+3)(n+3)(n-3)$.

Step 3. Subtract the fractions

Rewrite each rational expression as an equivalent rational expression with the LCD.

$\frac{4}{(n+3)(n+3)}-\frac{1}{(n+3)(n-3)}=\frac{4·(n-3)}{(n+3)(n+3)(n-3)}-\frac{1·(n+3)}{(n+3)(n-3)(n+3)}$

Now on simplifying the numerator we get

$\begin{array}{rcl}\frac{4·(n-3)}{(n+3)(n+3)(n-3)}-\frac{1·(n+3)}{(n+3)(n-3)(n+3)}& =& \frac{4n-12-(n+3)}{(n+3)(n+3)(n-3)}\\ & =& \frac{4n-12-n-3}{(n+3)(n+3)(n-3)}\\ & =& \frac{3n-15}{(n+3)(n+3)(n-3)}\end{array}$