Find the Holes in the Graph (-4x-16)/(x+4)

$\frac{- 4 x - 16}{x + 4}$

Factor $4$ out of $- 4 x - 16$ .

Factor $4$ out of $- 4 x$ .

$\frac{4 (- x) - 16}{x + 4}$

Factor $4$ out of $- 16$ .

$\frac{4 (- x) + 4 (- 4)}{x + 4}$

Factor $4$ out of $4 (- x) + 4 (- 4)$ .

$\frac{4 (- x - 4)}{x + 4}$

Cancel the common factor of $- x - 4$ and $x + 4$ .

Factor $- 1$ out of $- x$ .

$\frac{4 (- (x) - 4)}{x + 4}$

Rewrite $- 4$ as $- 1 (4)$ .

$\frac{4 (- (x) - 1 (4))}{x + 4}$

Factor $- 1$ out of $- (x) - 1 (4)$ .

$\frac{4 (- (x + 4))}{x + 4}$

Rewrite $- (x + 4)$ as $- 1 (x + 4)$ .

$\frac{4 (- 1 (x + 4))}{x + 4}$

Cancel the common factor.

$\frac{4 (- 1 (x + 4))}{x + 4}$

Divide $4 \cdot (- 1)$ by $1$ .

$4 \cdot (- 1)$

$4 (- 1)$

Multiply $4$ by $- 1$ .

$- 4$

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

$x + 4$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- 4$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Because the cancelled expression is constant with respect to $x$ , it is the $y$ coordinate for every hole.

$- 4$

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

$(- 4, - 4)$