Find the Holes in the Graph f(x)=(2x^2-8x-24)/(x^2-36)

$f (x) = \frac{2 x^{2} - 8 x - 24}{x^{2} - 36}$

Factor $2 x^{2} - 8 x - 24$ .

Factor $2$ out of $2 x^{2} - 8 x - 24$ .

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

$f (x) = \frac{2 (x^{2}) - 8 x - 24}{x^{2} - 36}$

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

$f (x) = \frac{2 (x^{2}) + 2 (- 4 x) - 24}{x^{2} - 36}$

Factor $2$ out of $- 24$ .

$f (x) = \frac{2 x^{2} + 2 (- 4 x) + 2 \cdot - 12}{x^{2} - 36}$

Factor $2$ out of $2 x^{2} + 2 (- 4 x)$ .

$f (x) = \frac{2 (x^{2} - 4 x) + 2 \cdot - 12}{x^{2} - 36}$

Factor $2$ out of $2 (x^{2} - 4 x) + 2 \cdot - 12$ .

$f (x) = \frac{2 (x^{2} - 4 x - 12)}{x^{2} - 36}$

Factor.

Factor $x^{2} - 4 x - 12$ 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 $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Write the factored form using these integers.

$f (x) = \frac{2 ((x - 6) (x + 2))}{x^{2} - 36}$

Remove unnecessary parentheses.

$f (x) = \frac{2 (x - 6) (x + 2)}{x^{2} - 36}$

Factor $x^{2} - 36$ .

Rewrite $36$ as $6^{2}$ .

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

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 $b = 6$ .

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

Cancel the common factor of $x - 6$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x - 6$

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

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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

Substitute $6$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Cancel the common factors.

Factor $2$ out of $6$ .

$\frac{2 (6 + 2)}{2 \cdot 3 + 6}$

Factor $2$ out of $6$ .

$\frac{2 (6 + 2)}{2 \cdot 3 + 2 \cdot 3}$

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

$\frac{2 (6 + 2)}{2 \cdot (3 + 3)}$

Cancel the common factor.

$\frac{2 (6 + 2)}{2 \cdot (3 + 3)}$

Rewrite the expression.

$\frac{6 + 2}{3 + 3}$

Add $6$ and $2$ .

$\frac{8}{3 + 3}$

Add $3$ and $3$ .

$\frac{8}{6}$

Cancel the common factor of $8$ and $6$ .

Factor $2$ out of $8$ .

$\frac{2 (4)}{6}$

Cancel the common factors.

Factor $2$ out of $6$ .

$\frac{2 \cdot 4}{2 \cdot 3}$

Cancel the common factor.

$\frac{2 \cdot 4}{2 \cdot 3}$

Rewrite the expression.

$\frac{4}{3}$

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

$(6, \frac{4}{3})$

Find the Holes in the Graph f(x)=(2x^2-8x-24)/(x^2-36)

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