Find the Holes in the Graph f(x)=(x^2-4)/(x^3-x^2-2x)

$f (x) = \frac{x^{2} - 4}{x^{3} - x^{2} - 2 x}$

Step 1

Factor $x^{2} - 4$ .

Rewrite $4$ as $2^{2}$ .

$f (x) = \frac{x^{2} - 2^{2}}{x^{3} - x^{2} - 2 x}$

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 $b = 2$ .

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

Step 2

Factor $x^{3} - x^{2} - 2 x$ .

Factor $x$ out of $x^{3} - x^{2} - 2 x$ .

Factor $x$ out of $x^{3}$ .

$f (x) = \frac{(x + 2) (x - 2)}{x \cdot x^{2} - x^{2} - 2 x}$

Factor $x$ out of $- x^{2}$ .

$f (x) = \frac{(x + 2) (x - 2)}{x \cdot x^{2} + x (- x) - 2 x}$

Factor $x$ out of $- 2 x$ .

$f (x) = \frac{(x + 2) (x - 2)}{x \cdot x^{2} + x (- x) + x \cdot - 2}$

Factor $x$ out of $x \cdot x^{2} + x (- x)$ .

$f (x) = \frac{(x + 2) (x - 2)}{x (x^{2} - x) + x \cdot - 2}$

Factor $x$ out of $x (x^{2} - x) + x \cdot - 2$ .

$f (x) = \frac{(x + 2) (x - 2)}{x (x^{2} - x - 2)}$

Factor.

Factor $x^{2} - x - 2$ 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 $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

$f (x) = \frac{(x + 2) (x - 2)}{x ((x - 2) (x + 1))}$

Remove unnecessary parentheses.

$f (x) = \frac{(x + 2) (x - 2)}{x (x - 2) (x + 1)}$

Step 3

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$f (x) = \frac{(x + 2) (x - 2)}{x (x - 2) (x + 1)}$

Rewrite the expression.

$f (x) = \frac{x + 2}{x (x + 1)}$

Step 4

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

$x - 2$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 2}{x (x + 1)}$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Substitute $2$ for $x$ in $\frac{x + 2}{x (x + 1)}$ and simplify.

Substitute $2$ for $x$ to find the $y$ coordinate of the hole.

$\frac{2 + 2}{2 (2 + 1)}$

Simplify.

Cancel the common factor of $2 + 2$ and $2$ .

Factor $2$ out of $2$ .

$\frac{2 \cdot 1 + 2}{2 (2 + 1)}$

Factor $2$ out of $2$ .

$\frac{2 \cdot 1 + 2 \cdot 1}{2 (2 + 1)}$

Factor $2$ out of $2 \cdot 1 + 2 \cdot 1$ .

$\frac{2 \cdot (1 + 1)}{2 (2 + 1)}$

Cancel the common factors.

Cancel the common factor.

$\frac{2 \cdot (1 + 1)}{2 (2 + 1)}$

Rewrite the expression.

$\frac{1 + 1}{2 + 1}$

Add $1$ and $1$ .

$\frac{2}{2 + 1}$

Add $2$ and $1$ .

$\frac{2}{3}$

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

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

Step 6

Find the Holes in the Graph f(x)=(x^2-4)/(x^3-x^2-2x)

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