# What are the excluded values for (5x+1)/(x^2-1)?

Apr 4, 2017

See the entire solution process below:

#### Explanation:

Because we cannot divide by $0$ the excluded values are:

${x}^{2} - 1 \ne 0$

We can factor ${x}^{2} - 1$ using the rule:

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

Letting ${a}^{2} = {x}^{2}$, $a = x$, ${b}^{2} = 1$ and $b = 1$ and substituting gives:

$\left(x + 1\right) \left(x - 1\right) \ne 0$

Now, solve each term for $0$ to find the excluded values of $x$:

Solution 1)

$x + 1 = 0$

$x + 1 - \textcolor{red}{1} = 0 - \textcolor{red}{1}$

$x + 0 = - 1$

$x = - 1$

Solution 2)

$x - 1 = 0$

$x - 1 + \textcolor{red}{1} = 0 + \textcolor{red}{1}$

$x - 0 = 1$

$x = 1$

The excluded values are: $x = - 1$ and $x = 1$