Algebra Examples

\frac{x^{2} - 1}{x - 1}

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

Step 1

Factor

x^{2} - 1

$x^{2} - 1$ .

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Step 1.1

Rewrite

1

$1$ as

1^{2}

$1^{2}$ .

\frac{x^{2} - 1^{2}}{x - 1}

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

Step 1.2

Since both terms are perfect squares, factor using the difference of squares formula,

a^{2} - b^{2} = (a + b) (a - b)

$a^{2} - b^{2} = (a + b) (a - b)$ where

a = x

$a = x$ and

b = 1

$b = 1$ .

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

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

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

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

Step 2

Cancel the common factor of

x - 1

$x - 1$ .

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Step 2.1

Cancel the common factor.

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

Step 2.2

Divide

$x + 1$ by

$1$ .

$x + 1$

Step 3

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

$x - 1$

Step 4

To find the coordinates of the holes, set each factor that was cancelled equal to

$0$ , solve, and substitute back in to

$x + 1$ .

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Step 4.1

Set

$x - 1$ equal to

$0$ .

$x - 1 = 0$

Step 4.2

Add

$1$ to both sides of the equation.

$x = 1$

Step 4.3

Substitute

$1$ for

$x$ in

$x + 1$ and simplify.

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Step 4.3.1

Substitute

$1$ for

$x$ to find the

$y$ coordinate of the hole.

$1 + 1$

Step 4.3.2

Add

$1$ and

$1$ .

$2$

Step 4.4

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

$0$ .

$(1, 2)$

Step 5