# How do you simplify (3x)/(x^2-x-12) - (x-1)/(x^2+6x+9) + (x-6)/(2x+6) ?

Jun 8, 2016

$\frac{3 x}{{x}^{2} - x - 12} - \frac{x - 1}{{x}^{2} + 6 x + 9} + \frac{x - 6}{2 x + 6} = \frac{{x}^{3} - 3 {x}^{2} + 22 x + 64}{2 \left(x + 3\right) \left(x + 3\right) \left(x - 4\right)}$

#### Explanation:

For simplifying this, we first need to factorize denominators (as numerators are already in simplest terms).

${x}^{2} - x - 12 = {x}^{2} - 4 x + 3 x - 12 = x \left(x - 4\right) + 3 \left(x - 4\right) = \left(x + 3\right) \left(x - 4\right)$

${x}^{2} + 6 x 9 = {x}^{2} + 3 x + 3 x + 9 = x \left(x + 3\right) + 3 \left(x + 3\right) = \left(x + 3\right) \left(x + 3\right) = {\left(x + 3\right)}^{2}$

$2 x + 6 = 2 \left(x + 3\right)$

And LCD of three denominators is $2 \left(x + 3\right) \left(x + 3\right) \left(x - 4\right)$

Hence $\frac{3 x}{{x}^{2} - x - 12} - \frac{x - 1}{{x}^{2} + 6 x + 9} + \frac{x - 6}{2 x + 6}$

= $\frac{3 x}{\left(x + 3\right) \left(x - 4\right)} - \frac{x - 1}{x + 3} ^ 2 + \frac{x - 6}{2 \left(x + 3\right)}$

= $\frac{3 x \times 2 \times \left(x + 3\right) - \left(x - 1\right) \times 2 \left(x - 4\right) + \left(x - 6\right) \left(x + 3\right) \left(x - 4\right)}{2 \left(x + 3\right) \left(x + 3\right) \left(x - 4\right)}$

= $\frac{\left(6 {x}^{2} + 18 x\right) - 2 \left({x}^{2} - 5 x + 4\right) + \left(x - 6\right) \left({x}^{2} - x - 12\right)}{2 \left(x + 3\right) \left(x + 3\right) \left(x - 4\right)}$

= $\frac{\left(6 {x}^{2} + 18 x\right) - 2 \left({x}^{2} - 5 x + 4\right) + \left({x}^{3} - {x}^{2} - 12 x - 6 {x}^{2} + 6 x + 72\right)}{2 \left(x + 3\right) \left(x + 3\right) \left(x - 4\right)}$

= $\frac{{x}^{3} - 3 {x}^{2} + 22 x + 64}{2 \left(x + 3\right) \left(x + 3\right) \left(x - 4\right)}$

Jun 8, 2016

$\frac{{x}^{3} - 3 {x}^{2} + 22 x + 64}{2 \left(x - 4\right) {\left(x + 3\right)}^{2}}$

#### Explanation:

Look for common factors. So we need to investigate factorisation of the denominators.

${x}^{2} - x - 12 = \left(x + 3\right) \left(x - 4\right)$

${x}^{2} + 6 x + 9 = \left(x + 3\right) \left(x + 3\right)$

$2 x + 6 = 2 \left(x + 3\right)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Factor out $\left(x + 3\right)$ from the denominators giving:

$\frac{1}{x + 3} \left[\textcolor{w h i t e}{.} \frac{3 x}{x - 4} - \frac{x - 1}{x + 3} + \frac{x - 6}{2} \textcolor{w h i t e}{.}\right] \textcolor{red}{\text{... Eqn (1)}}$

Using a common denominator of $2 \left(x - 4\right) \left(x + 3\right)$

'.................................................................................................................
Consider $\frac{3 x}{x - 4} \to \frac{3 x \times 2 \times \left(x + 3\right)}{2 \left(x - 4\right) \left(x + 3\right)} = \frac{6 {x}^{2} + 18 x}{2 \left(x - 4\right) \left(x + 3\right)}$
,...................................................................................................................

Consider $- \frac{x - 1}{x + 3} \to \frac{\left(x - 1\right) \times 2 \times \left(x - 4\right)}{2 \left(x - 4\right) \left(x + 3\right)} = - \frac{2 {x}^{2} - 10 x + 8}{2 \left(x - 4\right) \left(x + 3\right)}$ .
'.....................................................................................................................

Consider $\frac{x - 6}{2} \to \frac{\left(x - 6\right) \left(x - 4\right) \left(x + 3\right)}{2 \left(x - 4\right) \left(x + 3\right)}$

=(x^3-7x^2-6x+72)/(2(x-4)(x+3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting it all together}}$

$\frac{1}{x + 3} \left[\frac{{x}^{3} - 3 {x}^{2} + 22 x + 64}{2 \left(x - 4\right) \left(x + 3\right)}\right]$

$\frac{{x}^{3} - 3 {x}^{2} + 22 x + 64}{2 \left(x + 3\right) \left(x - 4\right) \left(x + 3\right)}$

Factoring this further

$\frac{\left(x + 2\right) \left({x}^{2} - 5 x + 32\right)}{2 \left(x - 4\right) {\left(x + 3\right)}^{2}} \leftarrow \text{ don't think this helps much}$