Find the Holes in the Graph F(x)=(x^2-9)/(x^2+7x+12)

$F (x) = \frac{x^{2} - 9}{x^{2} + 7 x + 12}$

Factor $x^{2} - 9$ .

Rewrite $9$ as $3^{2}$ .

$F (x) = \frac{x^{2} - 3^{2}}{x^{2} + 7 x + 12}$

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 $b = 3$ .

$F (x) = \frac{(x + 3) (x - 3)}{x^{2} + 7 x + 12}$

Factor $x^{2} + 7 x + 12$ 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 $7$ .

$3, 4$

Write the factored form using these integers.

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

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x + 3$

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

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Substitute $- 3$ for $x$ in $\frac{x - 3}{x + 4}$ and simplify.

Substitute $- 3$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Subtract $3$ from $- 3$ .

$\frac{- 6}{- 3 + 4}$

Add $- 3$ and $4$ .

$\frac{- 6}{1}$

Divide $- 6$ by $1$ .

$- 6$

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

$(- 3, - 6)$

Find the Holes in the Graph F(x)=(x^2-9)/(x^2+7x+12)

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