Find the Holes in the Graph (x^2-2x-3)/(-4x+12)

$\frac{x^{2} - 2 x - 3}{- 4 x + 12}$

Step 1

Factor $x^{2} - 2 x - 3$ 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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$\frac{(x - 3) (x + 1)}{- 4 x + 12}$

Step 2

Factor $4$ out of $- 4 x + 12$ .

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

$\frac{(x - 3) (x + 1)}{4 (- x) + 12}$

Factor $4$ out of $12$ .

$\frac{(x - 3) (x + 1)}{4 (- x) + 4 (3)}$

Factor $4$ out of $4 (- x) + 4 (3)$ .

$\frac{(x - 3) (x + 1)}{4 (- x + 3)}$

Step 3

Cancel the common factor of $x - 3$ and $- x + 3$ .

Factor $- 1$ out of $x$ .

$\frac{(- 1 (- x) - 3) (x + 1)}{4 (- x + 3)}$

Rewrite $- 3$ as $- 1 (3)$ .

$\frac{(- 1 (- x) - 1 (3)) (x + 1)}{4 (- x + 3)}$

Factor $- 1$ out of $- 1 (- x) - 1 (3)$ .

$\frac{- 1 (- x + 3) (x + 1)}{4 (- x + 3)}$

Cancel the common factor.

$\frac{- 1 (- x + 3) (x + 1)}{4 (- x + 3)}$

Rewrite the expression.

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

Step 4

Move the negative in front of the fraction.

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

Step 5

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

$- x + 3$

Step 6

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{x + 1}{4}$ .

Set $- x + 3$ equal to $0$ .

$- x + 3 = 0$

Solve $- x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$- x = - 3$

Divide each term in $- x = - 3$ by $- 1$ and simplify.

Divide each term in $- x = - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

Substitute $3$ for $x$ in $- \frac{x + 1}{4}$ and simplify.

Substitute $3$ for $x$ to find the $y$ coordinate of the hole.

$- \frac{3 + 1}{4}$

Simplify.

Add $3$ and $1$ .

$- \frac{4}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$- \frac{4}{4}$

Rewrite the expression.

$- 1 \cdot 1$

Multiply $- 1$ by $1$ .

$- 1$

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

$(3, - 1)$

Step 7

Find the Holes in the Graph (x^2-2x-3)/(-4x+12)

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