Find the Holes in the Graph f(x)=(3x^2-6x-24)/(2x^2-8)

$f (x) = \frac{3 x^{2} - 6 x - 24}{2 x^{2} - 8}$

Step 1

Factor $3 x^{2} - 6 x - 24$ .

Factor $3$ out of $3 x^{2} - 6 x - 24$ .

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

$f (x) = \frac{3 (x^{2}) - 6 x - 24}{2 x^{2} - 8}$

Factor $3$ out of $- 6 x$ .

$f (x) = \frac{3 (x^{2}) + 3 (- 2 x) - 24}{2 x^{2} - 8}$

Factor $3$ out of $- 24$ .

$f (x) = \frac{3 x^{2} + 3 (- 2 x) + 3 \cdot - 8}{2 x^{2} - 8}$

Factor $3$ out of $3 x^{2} + 3 (- 2 x)$ .

$f (x) = \frac{3 (x^{2} - 2 x) + 3 \cdot - 8}{2 x^{2} - 8}$

Factor $3$ out of $3 (x^{2} - 2 x) + 3 \cdot - 8$ .

$f (x) = \frac{3 (x^{2} - 2 x - 8)}{2 x^{2} - 8}$

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.

$f (x) = \frac{3 ((x - 4) (x + 2))}{2 x^{2} - 8}$

Remove unnecessary parentheses.

$f (x) = \frac{3 (x - 4) (x + 2)}{2 x^{2} - 8}$

Step 2

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

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

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

$f (x) = \frac{3 (x - 4) (x + 2)}{2 (x^{2}) - 8}$

Factor $2$ out of $- 8$ .

$f (x) = \frac{3 (x - 4) (x + 2)}{2 x^{2} + 2 \cdot - 4}$

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

$f (x) = \frac{3 (x - 4) (x + 2)}{2 (x^{2} - 4)}$

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

$f (x) = \frac{3 (x - 4) (x + 2)}{2 (x^{2} - 2^{2})}$

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 $b = 2$ .

$f (x) = \frac{3 (x - 4) (x + 2)}{2 ((x + 2) (x - 2))}$

Remove unnecessary parentheses.

$f (x) = \frac{3 (x - 4) (x + 2)}{2 (x + 2) (x - 2)}$

Step 3

Cancel the common factor of $x + 2$ .

Cancel the common factor.

$f (x) = \frac{3 (x - 4) (x + 2)}{2 (x + 2) (x - 2)}$

Rewrite the expression.

$f (x) = \frac{3 (x - 4)}{2 (x - 2)}$

Step 4

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

$x + 2$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{3 (x - 4)}{2 (x - 2)}$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Substitute $- 2$ for $x$ in $\frac{3 (x - 4)}{2 (x - 2)}$ and simplify.

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

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

Simplify.

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

Factor $2$ out of $3 (- 2 - 4)$ .

$\frac{2 (3 (- 1 - 2))}{2 (- 2 - 2)}$

Cancel the common factors.

Cancel the common factor.

$\frac{2 (3 (- 1 - 2))}{2 (- 2 - 2)}$

Rewrite the expression.

$\frac{3 (- 1 - 2)}{- 2 - 2}$

Subtract $2$ from $- 1$ .

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

Subtract $2$ from $- 2$ .

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

Multiply $3$ by $- 3$ .

$\frac{- 9}{- 4}$

Dividing two negative values results in a positive value.

$\frac{9}{4}$

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

$(- 2, \frac{9}{4})$

Step 6

Find the Holes in the Graph f(x)=(3x^2-6x-24)/(2x^2-8)

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