Find the Holes in the Graph f(x)=(x^2-16)/(2x+8)

$f (x) = \frac{x^{2} - 16}{2 x + 8}$

Factor $x^{2} - 16$ .

Rewrite $16$ as $4^{2}$ .

$f (x) = \frac{x^{2} - 4^{2}}{2 x + 8}$

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 $b = 4$ .

$f (x) = \frac{(x + 4) (x - 4)}{2 x + 8}$

Factor $2$ out of $2 x + 8$ .

Factor $2$ out of $2 x$ .

$f (x) = \frac{(x + 4) (x - 4)}{2 (x) + 8}$

Factor $2$ out of $8$ .

$f (x) = \frac{(x + 4) (x - 4)}{2 x + 2 \cdot 4}$

Factor $2$ out of $2 x + 2 \cdot 4$ .

$f (x) = \frac{(x + 4) (x - 4)}{2 (x + 4)}$

Cancel the common factor of $x + 4$ .

Cancel the common factor.

$f (x) = \frac{(x + 4) (x - 4)}{2 (x + 4)}$

Rewrite the expression.

$f (x) = \frac{x - 4}{2}$

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

$x + 4$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x - 4}{2}$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Substitute $- 4$ for $x$ in $\frac{x - 4}{2}$ and simplify.

Substitute $- 4$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Cancel the common factor of $- 4 - 4$ and $2$ .

Factor $2$ out of $- 4$ .

$\frac{2 \cdot - 2 - 4}{2}$

Factor $2$ out of $- 4$ .

$\frac{2 \cdot - 2 + 2 \cdot - 2}{2}$

Factor $2$ out of $2 \cdot - 2 + 2 \cdot - 2$ .

$\frac{2 \cdot (- 2 - 2)}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 2 - 2$ by $1$ .

$- 2 - 2$

Subtract $2$ from $- 2$ .

$- 4$

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

$(- 4, - 4)$

Find the Holes in the Graph f(x)=(x^2-16)/(2x+8)

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