# How do you simplify and restricted value of (x^3 - 2x + 3)/( x^2 + 12x + 32)?

May 19, 2017

$\frac{{x}^{3} - 2 x + 3}{{\left(x + 6\right)}^{2}}$; $x \ne - 6$

#### Explanation:

We have: $\frac{{x}^{3} - 2 x + 3}{{x}^{2} + 12 x + 32}$

Let's factorise the denominator using the "middle-term break":

$= \frac{{x}^{3} - 2 x + 3}{{x}^{2} + 6 x + 6 x + 32}$

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

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

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

Now let's determine the restricted values of $x$.

The denominator of the fraction can never equal to zero:

$R i g h t a r r o w {\left(x + 6\right)}^{2} \ne 0$

$R i g h t a r r o w x + 6 \ne 0$

$R i g h t a r r o w x \ne - 6$

$\therefore \frac{{x}^{3} - 2 x + 3}{{x}^{2} + 12 x + 32} = \frac{{x}^{3} - 2 x + 3}{{\left(x + 6\right)}^{2}}$; $x \ne - 6$