Find the Holes in the Graph (x^2-1)/(x-1)

$\frac{x^{2} - 1}{x - 1}$

Factor $x^{2} - 1$ .

Rewrite $1$ as $1^{2}$ .

$\frac{x^{2} - 1^{2}}{x - 1}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$\frac{(x + 1) (x - 1)}{x - 1}$

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$\frac{(x + 1) (x - 1)}{x - 1}$

Divide $x + 1$ by $1$ .

$x + 1$

To find the holes in the graph, look at the denominator factors that were cancelled.

$x - 1$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $x + 1$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Substitute $1$ for $x$ in $x + 1$ and simplify.

Substitute $1$ for $x$ to find the $y$ coordinate of the hole.

$1 + 1$

Add $1$ and $1$ .

$2$

The holes in the graph are the points where any of the cancelled factors are equal to $0$ .

$(1, 2)$