Find the Holes in the Graph f(x)=(x^2+6x+9)/(x+3)

$f (x) = \frac{x^{2} + 6 x + 9}{x + 3}$

Factor using the perfect square rule.

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

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 3$ .

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

Cancel the common factor of ${(x + 3)}^{2}$ and $x + 3$ .

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

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

Cancel the common factors.

Multiply by $1$ .

$f (x) = \frac{(x + 3) (x + 3)}{(x + 3) \cdot 1}$

Cancel the common factor.

$f (x) = \frac{(x + 3) (x + 3)}{(x + 3) \cdot 1}$

Rewrite the expression.

$f (x) = \frac{x + 3}{1}$

Divide $x + 3$ by $1$ .

$f (x) = x + 3$

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 $x + 3$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Substitute $- 3$ for $x$ in $x + 3$ and simplify.

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

$- 3 + 3$

Add $- 3$ and $3$ .

$0$

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

$(- 3, 0)$