Find the Holes in the Graph f(x)=(2x^2+16x)/(2x+16)

$f (x) = \frac{2 x^{2} + 16 x}{2 x + 16}$

Factor $2 x$ out of $2 x^{2} + 16 x$ .

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

$f (x) = \frac{2 x (x) + 16 x}{2 x + 16}$

Factor $2 x$ out of $16 x$ .

$f (x) = \frac{2 x (x) + 2 x (8)}{2 x + 16}$

Factor $2 x$ out of $2 x (x) + 2 x (8)$ .

$f (x) = \frac{2 x (x + 8)}{2 x + 16}$

Factor $2$ out of $2 x + 16$ .

Factor $2$ out of $2 x$ .

$f (x) = \frac{2 x (x + 8)}{2 (x) + 16}$

Factor $2$ out of $16$ .

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

Factor $2$ out of $2 x + 2 \cdot 8$ .

$f (x) = \frac{2 x (x + 8)}{2 (x + 8)}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (x) = \frac{2 x (x + 8)}{2 (x + 8)}$

Rewrite the expression.

$f (x) = \frac{x (x + 8)}{x + 8}$

Cancel the common factor of $x + 8$ .

Cancel the common factor.

$f (x) = \frac{x (x + 8)}{x + 8}$

Divide $x$ by $1$ .

$f (x) = x$

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

$x + 8$

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

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

Substitute $- 8$ for $x$ to find the $y$ coordinate of the hole.

$- 8$

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

$(- 8, - 8)$