Find the Holes in the Graph (4x+8)/(x^2+2x)

$\frac{4 x + 8}{x^{2} + 2 x}$

Step 1

Factor $4$ out of $4 x + 8$ .

Factor $4$ out of $4 x$ .

$\frac{4 (x) + 8}{x^{2} + 2 x}$

Factor $4$ out of $8$ .

$\frac{4 x + 4 \cdot 2}{x^{2} + 2 x}$

Factor $4$ out of $4 x + 4 \cdot 2$ .

$\frac{4 (x + 2)}{x^{2} + 2 x}$

Step 2

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

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

$\frac{4 (x + 2)}{x \cdot x + 2 x}$

Factor $x$ out of $2 x$ .

$\frac{4 (x + 2)}{x \cdot x + x \cdot 2}$

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

$\frac{4 (x + 2)}{x (x + 2)}$

Step 3

Cancel the common factor of $x + 2$ .

Cancel the common factor.

$\frac{4 (x + 2)}{x (x + 2)}$

Rewrite the expression.

$\frac{4}{x}$

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 $\frac{4}{x}$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Substitute $- 2$ for $x$ in $\frac{4}{x}$ and simplify.

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

$\frac{4}{- 2}$

Divide $4$ by $- 2$ .

$- 2$

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

$(- 2, - 2)$

Step 6