# How do you simplify (8n-3)/(n^2+8n+12)-(5n-9)/(n^2+8n+12)?

Aug 8, 2018

$\frac{8 n - 3}{{n}^{2} + 8 n + 12} - \frac{5 n - 9}{{n}^{2} + 8 n + 12} = \frac{3}{n + 6}$

#### Explanation:

Let ,

$P = \frac{8 n - 3}{{n}^{2} + 8 n + 12} - \frac{5 n - 9}{{n}^{2} + 8 n + 12}$

$P = \frac{\left(8 n - 3\right) - \left(5 n - 9\right)}{{n}^{2} + 8 n + 12} \leftarrow \ldots \ldots \ldots \text{Combine fractions}$

$P = \frac{8 n - 3 - 5 n + 9}{{n}^{2} + 8 n + 12}$

$P = \frac{3 n + 6}{{n}^{2} + 8 n + 12} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \to \left(1\right)$

Now , we obtain factors:

$3 n + 6 = 3 \left(n + 2\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \to \left(2\right)$

Here, $\left(6\right) + \left(2\right) = 8 \mathmr{and} \left(6\right) \times \left(2\right) = 12$

So ,

${n}^{2} + \textcolor{red}{8 n} + 12 = {n}^{2} \textcolor{red}{+ 6 n + 2 n} + 12$

${n}^{2} + 8 n + 12 = n \left(n + 6\right) + 2 \left(n + 6\right)$

${n}^{2} + 8 n + 12 = \left(n + 6\right) \left(n + 2\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \to \left(3\right)$

From $\left(1\right) , \left(2\right) \mathmr{and} \left(3\right)$

$P = \frac{3 \cancel{\left(n + 2\right)}}{\left(n + 6\right) \cancel{\left(n + 2\right)}}$

$\therefore P = \frac{3}{n + 6}$