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

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

Step 1

Factor $2 x$ out of $- 2 x^{2} - 4 x$ .

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

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

Factor $2 x$ out of $- 4 x$ .

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

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

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

Step 2

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

Factor $4$ out of $4 x$ .

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

Factor $4$ out of $8$ .

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

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

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

Step 3

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2 x (- x - 2)$ .

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

Cancel the common factors.

Factor $2$ out of $4 (x + 2)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Cancel the common factor of $- x - 2$ and $x + 2$ .

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

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

Rewrite $- 2$ as $- 1 (2)$ .

$\frac{x (- (x) - 1 (2))}{2 (x + 2)}$

Factor $- 1$ out of $- (x) - 1 (2)$ .

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

Rewrite $- (x + 2)$ as $- 1 (x + 2)$ .

$\frac{x (- 1 (x + 2))}{2 (x + 2)}$

Cancel the common factor.

$\frac{x (- 1 (x + 2))}{2 (x + 2)}$

Rewrite the expression.

$\frac{x \cdot (- 1)}{2}$

Step 5

Move $- 1$ to the left of $x$ .

$\frac{- 1 \cdot x}{2}$

Step 6

Move the negative in front of the fraction.

$- \frac{x}{2}$

Step 7

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

$x + 2$

Step 8

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{x}{2}$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

$- \frac{- 2}{2}$

Simplify.

Divide $- 2$ by $2$ .

$- - 1$

Multiply $- 1$ by $- 1$ .

$1$

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

$(- 2, 1)$

Step 9

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

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