Find the Holes in the Graph g(x)=((x-2)(x+6))/((x+2)(x-2))

$g (x) = \frac{(x - 2) (x + 6)}{(x + 2) (x - 2)}$

Step 1

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$g (x) = \frac{(x - 2) (x + 6)}{(x + 2) (x - 2)}$

Rewrite the expression.

$g (x) = \frac{x + 6}{x + 2}$

Step 2

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

$x - 2$

Step 3

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 6}{x + 2}$ .

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Substitute $2$ for $x$ in $\frac{x + 6}{x + 2}$ and simplify.

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

$\frac{2 + 6}{2 + 2}$

Simplify.

Cancel the common factor of $2 + 6$ and $2 + 2$ .

Factor $2$ out of $2$ .

$\frac{2 \cdot 1 + 6}{2 + 2}$

Factor $2$ out of $6$ .

$\frac{2 \cdot 1 + 2 \cdot 3}{2 + 2}$

Factor $2$ out of $2 \cdot 1 + 2 \cdot 3$ .

$\frac{2 \cdot (1 + 3)}{2 + 2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$\frac{2 \cdot (1 + 3)}{2 \cdot 1 + 2}$

Factor $2$ out of $2$ .

$\frac{2 \cdot (1 + 3)}{2 \cdot 1 + 2 (1)}$

Factor $2$ out of $2 \cdot 1 + 2 (1)$ .

$\frac{2 \cdot (1 + 3)}{2 \cdot (1 + 1)}$

Cancel the common factor.

$\frac{2 \cdot (1 + 3)}{2 \cdot (1 + 1)}$

Rewrite the expression.

$\frac{1 + 3}{1 + 1}$

Add $1$ and $3$ .

$\frac{4}{1 + 1}$

Add $1$ and $1$ .

$\frac{4}{2}$

Divide $4$ by $2$ .

$2$

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

$(2, 2)$

Step 4

Find the Holes in the Graph g(x)=((x-2)(x+6))/((x+2)(x-2))

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