Algebra Examples

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

Step 1

Factor $x^{2} - 2 x - 3$ using the AC method.

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Step 1.1

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 $- 2$ .

$- 3, 1$

Step 1.2

Write the factored form using these integers.

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

Step 2

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

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Step 2.1

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

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

Step 2.2

Factor $4$ out of $12$ .

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

Step 2.3

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 $- x + 3$ .

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Step 3.1

Factor $- 1$ out of $x$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Cancel the common factor.

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

Step 3.5

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

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Step 6.1

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

$- x + 3 = 0$

Step 6.2

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

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Step 6.2.1

Subtract $3$ from both sides of the equation.

$- x = - 3$

Step 6.2.2

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

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Step 6.2.2.1

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

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

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 6.2.2.2.2

Divide $x$ by $1$ .

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

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Divide $- 3$ by $- 1$ .

$x = 3$

Step 6.3

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

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Step 6.3.1

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

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

Step 6.3.2

Simplify.

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Step 6.3.2.1

Add $3$ and $1$ .

$- \frac{4}{4}$

Step 6.3.2.2

Cancel the common factor of $4$ .

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Step 6.3.2.2.1

Cancel the common factor.

$- \frac{4}{4}$

Step 6.3.2.2.2

Rewrite the expression.

$- 1 \cdot 1$

Step 6.3.2.3

Multiply $- 1$ by $1$ .

$- 1$

Step 6.4

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

$(3, - 1)$

Step 7