Find the Holes in the Graph (x^2-4x-5)/(4x+4)

$\frac{x^{2} - 4 x - 5}{4 x + 4}$

Step 1

Factor $x^{2} - 4 x - 5$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $- 4$ .

$- 5, 1$

Write the factored form using these integers.

$\frac{(x - 5) (x + 1)}{4 x + 4}$

Step 2

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

Factor $4$ out of $4 x$ .

$\frac{(x - 5) (x + 1)}{4 (x) + 4}$

Factor $4$ out of $4$ .

$\frac{(x - 5) (x + 1)}{4 (x) + 4 (1)}$

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

$\frac{(x - 5) (x + 1)}{4 (x + 1)}$

Step 3

Cancel the common factor of $x + 1$ .

Cancel the common factor.

$\frac{(x - 5) (x + 1)}{4 (x + 1)}$

Rewrite the expression.

$\frac{x - 5}{4}$

Step 4

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

$x + 1$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x - 5}{4}$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Substitute $- 1$ for $x$ in $\frac{x - 5}{4}$ and simplify.

Substitute $- 1$ for $x$ to find the $y$ coordinate of the hole.

$\frac{- 1 - 5}{4}$

Simplify.

Subtract $5$ from $- 1$ .

$\frac{- 6}{4}$

Cancel the common factor of $- 6$ and $4$ .

Factor $2$ out of $- 6$ .

$\frac{2 (- 3)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$\frac{2 \cdot - 3}{2 \cdot 2}$

Cancel the common factor.

$\frac{2 \cdot - 3}{2 \cdot 2}$

Rewrite the expression.

$\frac{- 3}{2}$

Move the negative in front of the fraction.

$- \frac{3}{2}$

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

$(- 1, - \frac{3}{2})$

Step 6

Find the Holes in the Graph (x^2-4x-5)/(4x+4)

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