Find the Holes in the Graph f(x)=(x^2-4)/(2x^2+10x+12)

$f (x) = \frac{x^{2} - 4}{2 x^{2} + 10 x + 12}$

Step 1

Factor $x^{2} - 4$ .

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

$f (x) = \frac{x^{2} - 2^{2}}{2 x^{2} + 10 x + 12}$

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 $b = 2$ .

$f (x) = \frac{(x + 2) (x - 2)}{2 x^{2} + 10 x + 12}$

Step 2

Factor $2 x^{2} + 10 x + 12$ .

Factor $2$ out of $2 x^{2} + 10 x + 12$ .

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

$f (x) = \frac{(x + 2) (x - 2)}{2 (x^{2}) + 10 x + 12}$

Factor $2$ out of $10 x$ .

$f (x) = \frac{(x + 2) (x - 2)}{2 (x^{2}) + 2 (5 x) + 12}$

Factor $2$ out of $12$ .

$f (x) = \frac{(x + 2) (x - 2)}{2 x^{2} + 2 (5 x) + 2 \cdot 6}$

Factor $2$ out of $2 x^{2} + 2 (5 x)$ .

$f (x) = \frac{(x + 2) (x - 2)}{2 (x^{2} + 5 x) + 2 \cdot 6}$

Factor $2$ out of $2 (x^{2} + 5 x) + 2 \cdot 6$ .

$f (x) = \frac{(x + 2) (x - 2)}{2 (x^{2} + 5 x + 6)}$

Factor.

Factor $x^{2} + 5 x + 6$ 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 $6$ and whose sum is $5$ .

$2, 3$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 3

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{x - 2}{2 (x + 3)}$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

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

Simplify.

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

Factor $2$ out of $- 2$ .

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

Factor $2$ out of $- 2$ .

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

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

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

$\frac{- 1 - 1}{- 2 + 3}$

Subtract $1$ from $- 1$ .

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

Add $- 2$ and $3$ .

$\frac{- 2}{1}$

Divide $- 2$ by $1$ .

$- 2$

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

$(- 2, - 2)$

Step 6

Find the Holes in the Graph f(x)=(x^2-4)/(2x^2+10x+12)

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