# How do you simplify (x^2+4x-32)/(x+8) and what are the ecluded values fot he variables?

May 29, 2017

#### Answer:

$x - 4$; $x \ne - 8$

#### Explanation:

We have: $\frac{{x}^{2} + 4 x - 32}{x + 8}$

Let's begin by factorising the numerator using the "middle-term break":

$= \frac{{x}^{2} + 8 x - 4 x - 32}{x + 8}$

$= \frac{x \left(x + 8\right) - 4 \left(x + 8\right)}{x + 8}$

$= \frac{\left(x + 8\right) \left(x - 4\right)}{x + 8}$

We can now cancel the $x + 8$ term:

$= x - 4$

Now, let's determine the excluded value of the variable $x$.

The denominator of a fraction can never equal to zero.

Let's set the denominator of the original fraction equal to zero and solve for $x$:

$R i g h t a r r o w x + 8 = 0$

$\therefore x = - 8$

So the excluded value of $x$ is $- 8$.

Therefore, $\frac{{x}^{2} + 4 x - 32}{x + 8} = x - 4$; $x \ne - 8$.