Find the Holes in the Graph f(x)=(3x^2-108)/(x^2+2x-24)

$f (x) = \frac{3 x^{2} - 108}{x^{2} + 2 x - 24}$

Factor $3 x^{2} - 108$ .

Factor $3$ out of $3 x^{2} - 108$ .

Factor $3$ out of $3 x^{2}$ .

$f (x) = \frac{3 (x^{2}) - 108}{x^{2} + 2 x - 24}$

Factor $3$ out of $- 108$ .

$f (x) = \frac{3 x^{2} + 3 \cdot - 36}{x^{2} + 2 x - 24}$

Factor $3$ out of $3 x^{2} + 3 \cdot - 36$ .

$f (x) = \frac{3 (x^{2} - 36)}{x^{2} + 2 x - 24}$

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

$f (x) = \frac{3 (x^{2} - 6^{2})}{x^{2} + 2 x - 24}$

Factor.

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 $b = 6$ .

$f (x) = \frac{3 ((x + 6) (x - 6))}{x^{2} + 2 x - 24}$

Remove unnecessary parentheses.

$f (x) = \frac{3 (x + 6) (x - 6)}{x^{2} + 2 x - 24}$

Factor $x^{2} + 2 x - 24$ 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 $- 24$ and whose sum is $2$ .

$- 4, 6$

Write the factored form using these integers.

$f (x) = \frac{3 (x + 6) (x - 6)}{(x - 4) (x + 6)}$

Cancel the common factor of $x + 6$ .

Cancel the common factor.

$f (x) = \frac{3 (x + 6) (x - 6)}{(x - 4) (x + 6)}$

Rewrite the expression.

$f (x) = \frac{3 (x - 6)}{x - 4}$

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 $\frac{3 (x - 6)}{x - 4}$ .

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

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Substitute $- 6$ for $x$ in $\frac{3 (x - 6)}{x - 4}$ and simplify.

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

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

Simplify.

Cancel the common factor of $- 6 - 6$ and $- 6 - 4$ .

Factor $2$ out of $3 (- 6 - 6)$ .

$\frac{2 (3 (- 3 - 3))}{- 6 - 4}$

Cancel the common factors.

Factor $2$ out of $- 6$ .

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

Factor $2$ out of $- 4$ .

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

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

$\frac{2 (3 (- 3 - 3))}{2 \cdot (- 3 - 2)}$

Cancel the common factor.

$\frac{2 (3 (- 3 - 3))}{2 \cdot (- 3 - 2)}$

Rewrite the expression.

$\frac{3 (- 3 - 3)}{- 3 - 2}$

Subtract $3$ from $- 3$ .

$\frac{3 \cdot - 6}{- 3 - 2}$

Subtract $2$ from $- 3$ .

$\frac{3 \cdot - 6}{- 5}$

Multiply $3$ by $- 6$ .

$\frac{- 18}{- 5}$

Dividing two negative values results in a positive value.

$\frac{18}{5}$

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

$(- 6, \frac{18}{5})$

Find the Holes in the Graph f(x)=(3x^2-108)/(x^2+2x-24)

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