Question

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

Short answer

The simplified answer is$\frac{3\left(n-5\right)}{{\left(n+3\right)}^2\left(n-3\right)}$

Step-by-step solution

Step 1. Given information.

The following expression is:

$\frac{4}{n^2+6n+9}-\frac{1}{n^2-9}$

Step 2. Simplification of the expression.

$$\begin{gathered} n^2+6n+9={\left(n+3\right)}^2 \\ {n}^2-9=\left(n+3\right)\left(n-3\right) \end{gathered}$$

Factorise the denominators:

$\frac{4}{{\left(n+3\right)}^2}-\frac{1}{\left(n+3\right)\left(n-3\right)}$

Find out the LCD of the denominators:

The LCD of the denominators will be

${\left(n+3\right)}^2\left(n-3\right)$

Rewrite each term with the denominator equal to the LCD:

$\frac{4\left(n-3\right)}{{\left(n+3\right)}^2\left(n-3\right)}-\frac{1\left(n+3\right)}{{\left(n+3\right)}^2\left(n-3\right)}$

Simplify:

$$\begin{gathered} \frac{4n-12-n-3}{{\left(n+3\right)}^2\left(n-3\right)} \\ \frac{3n-15}{{\left(n+3\right)}^2\left(n-3\right)} \end{gathered}$$

Further simplify:

$\frac{3\left(n-5\right)}{{\left(n+3\right)}^2\left(n-3\right)}$