Find the Holes in the Graph (x^2-16)/(2x^2-4x-16)

$\frac{x^{2} - 16}{2 x^{2} - 4 x - 16}$

Step 1

Factor $x^{2} - 16$ .

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

$\frac{x^{2} - 4^{2}}{2 x^{2} - 4 x - 16}$

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 $b = 4$ .

$\frac{(x + 4) (x - 4)}{2 x^{2} - 4 x - 16}$

Step 2

Factor $2 x^{2} - 4 x - 16$ .

Factor $2$ out of $2 x^{2} - 4 x - 16$ .

Factor $2$ out of $2 x^{2}$ .

$\frac{(x + 4) (x - 4)}{2 (x^{2}) - 4 x - 16}$

Factor $2$ out of $- 4 x$ .

$\frac{(x + 4) (x - 4)}{2 (x^{2}) + 2 (- 2 x) - 16}$

Factor $2$ out of $- 16$ .

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

Factor $2$ out of $2 x^{2} + 2 (- 2 x)$ .

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

Factor $2$ out of $2 (x^{2} - 2 x) + 2 \cdot - 8$ .

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

Factor.

Factor $x^{2} - 2 x - 8$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $- 2$ .

$- 4, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 3

Cancel the common factor of $x - 4$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{x + 4}{2 (x + 2)}$

Step 4

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

$x - 4$

Step 5

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Substitute $4$ for $x$ in $\frac{x + 4}{2 (x + 2)}$ and simplify.

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

$\frac{4 + 4}{2 (4 + 2)}$

Simplify.

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

Factor $2$ out of $4$ .

$\frac{2 \cdot 2 + 4}{2 (4 + 2)}$

Factor $2$ out of $4$ .

$\frac{2 \cdot 2 + 2 \cdot 2}{2 (4 + 2)}$

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

$\frac{2 \cdot (2 + 2)}{2 (4 + 2)}$

Cancel the common factors.

Cancel the common factor.

$\frac{2 \cdot (2 + 2)}{2 (4 + 2)}$

Rewrite the expression.

$\frac{2 + 2}{4 + 2}$

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

Factor $2$ out of $2$ .

$\frac{2 \cdot 1 + 2}{4 + 2}$

Factor $2$ out of $2$ .

$\frac{2 \cdot 1 + 2 \cdot 1}{4 + 2}$

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

$\frac{2 \cdot (1 + 1)}{4 + 2}$

Cancel the common factors.

Factor $2$ out of $4$ .

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

Factor $2$ out of $2$ .

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Add $1$ and $1$ .

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

Add $2$ and $1$ .

$\frac{2}{3}$

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

$(4, \frac{2}{3})$

Step 6

Find the Holes in the Graph (x^2-16)/(2x^2-4x-16)

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