# How do you simplify (r+6)/(4^2-r-6) and state the restrictions?

Apr 7, 2015

I assume your question is $\frac{r + 6}{{r}^{2} - r - 6}$

Let's Factorise the $\textcolor{red}{D E N O M I N A T O R}$ first.

The Denominator is $\textcolor{red}{{r}^{2} - r - 6}$

We can use Splitting the Middle Term technique to factorise this.

It is in the form $a {x}^{2} + b x + c$ where $a = 1 , b = - 1 , c = - 6$

To split the middle term, we need to think of two numbers ${N}_{1} \mathmr{and} {N}_{2}$ such that:
${N}_{1} \cdot {N}_{2} = a \cdot c \mathmr{and} {N}_{1} + {N}_{2} = b$
${N}_{1} \cdot {N}_{2} = \left(1\right) \cdot \left(- 6\right) \mathmr{and} {N}_{1} + {N}_{2} = - 1$
${N}_{1} \cdot {N}_{2} = - 6 \mathmr{and} {N}_{1} + {N}_{2} = - 1$

After Trial and Error, we get ${N}_{1} = 2 \mathmr{and} {N}_{2} = - 3$
$\left(2\right) \cdot \left(- 3\right) = - 6$ and $\left(2\right) + \left(- 3\right) = - 1$

So we can write the denominator as
$\textcolor{red}{{r}^{2} + 2 r - 3 r - 6}$
$= r \cdot \left(r + 2\right) - 3 \cdot \left(r + 2\right)$
$= \left(r + 2\right) \cdot \left(r - 3\right)$

The Denominator can be written as $\textcolor{red}{\left(r - 2\right) \cdot \left(r + 3\right)}$

The expression we have been given is
$\frac{r + 6}{{r}^{2} - r - 6}$

After the denominator was factorised, the Expression can now be written as :

$\frac{\left(r + 6\right)}{\left(r + 2\right) \cdot \left(r - 3\right)}$

$\frac{r + 6}{{r}^{2} - r - 6}$ = $\frac{\left(r + 6\right)}{\left(r + 2\right) \cdot \left(r - 3\right)}$