Find the Holes in the Graph (-3x+3)/(x^2-1)

$\frac{- 3 x + 3}{x^{2} - 1}$

Factor $3$ out of $- 3 x + 3$ .

Factor $3$ out of $- 3 x$ .

$\frac{3 (- x) + 3}{x^{2} - 1}$

Factor $3$ out of $3$ .

$\frac{3 (- x) + 3 (1)}{x^{2} - 1}$

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

$\frac{3 (- x + 1)}{x^{2} - 1}$

Factor $x^{2} - 1$ .

Rewrite $1$ as $1^{2}$ .

$\frac{3 (- x + 1)}{x^{2} - 1^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Cancel the common factor of $- x + 1$ and $x - 1$ .

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

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

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

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

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Multiply $3$ by $- 1$ .

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

Multiply $3$ by $- 1$ .

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

Move the negative in front of the fraction.

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

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

$x - 1$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{3}{x + 1}$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

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

Add $1$ and $1$ .

$- \frac{3}{2}$

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

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

Find the Holes in the Graph (-3x+3)/(x^2-1)

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