Find the Holes in the Graph (x^2-4)/(-x-2)

$\frac{x^{2} - 4}{- x - 2}$

Factor $x^{2} - 4$ .

Rewrite $4$ as $2^{2}$ .

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

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 $b = 2$ .

$\frac{(x + 2) (x - 2)}{- x - 2}$

Cancel the common factor of $x + 2$ and $- x - 2$ .

Factor $- 1$ out of $x$ .

$\frac{(- 1 (- x) + 2) (x - 2)}{- x - 2}$

Rewrite $2$ as $- 1 (- 2)$ .

$\frac{(- 1 (- x) - 1 (- 2)) (x - 2)}{- x - 2}$

Factor $- 1$ out of $- 1 (- x) - 1 (- 2)$ .

$\frac{- 1 (- x - 2) (x - 2)}{- x - 2}$

Cancel the common factor.

$\frac{- 1 (- x - 2) (x - 2)}{- x - 2}$

Divide $- 1 (x - 2)$ by $1$ .

$- 1 (x - 2)$

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

$- x - 2$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- 1 (x - 2)$ .

Set $- x - 2$ equal to $0$ .

$- x - 2 = 0$

Solve $- x - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$- x = 2$

Divide each term in $- x = 2$ by $- 1$ and simplify.

Divide each term in $- x = 2$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $2$ by $- 1$ .

$x = - 2$

Substitute $- 2$ for $x$ in $- 1 (x - 2)$ and simplify.

Substitute $- 2$ for $x$ to find the $y$ coordinate of the hole.

$- 1 (- 2 - 2)$

Simplify.

Subtract $2$ from $- 2$ .

$- 1 \cdot - 4$

Multiply $- 1$ by $- 4$ .

$4$

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

$(- 2, 4)$

Find the Holes in the Graph (x^2-4)/(-x-2)

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