Find the Holes in the Graph (x^2-25)/(x^2-4x-45)

$\frac{x^{2} - 25}{x^{2} - 4 x - 45}$

Factor $x^{2} - 25$ .

Rewrite $25$ as $5^{2}$ .

$\frac{x^{2} - 5^{2}}{x^{2} - 4 x - 45}$

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 $b = 5$ .

$\frac{(x + 5) (x - 5)}{x^{2} - 4 x - 45}$

Factor $x^{2} - 4 x - 45$ 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 $- 45$ and whose sum is $- 4$ .

$- 9, 5$

Write the factored form using these integers.

$\frac{(x + 5) (x - 5)}{(x - 9) (x + 5)}$

Cancel the common factor of $x + 5$ .

Cancel the common factor.

$\frac{(x + 5) (x - 5)}{(x - 9) (x + 5)}$

Rewrite the expression.

$\frac{x - 5}{x - 9}$

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

$x + 5$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x - 5}{x - 9}$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Substitute $- 5$ for $x$ in $\frac{x - 5}{x - 9}$ and simplify.

Substitute $- 5$ for $x$ to find the $y$ coordinate of the hole.

$\frac{- 5 - 5}{- 5 - 9}$

Simplify.

Subtract $5$ from $- 5$ .

$\frac{- 10}{- 5 - 9}$

Subtract $9$ from $- 5$ .

$\frac{- 10}{- 14}$

Cancel the common factor of $- 10$ and $- 14$ .

Factor $- 2$ out of $- 10$ .

$\frac{- 2 (5)}{- 14}$

Cancel the common factors.

Factor $- 2$ out of $- 14$ .

$\frac{- 2 \cdot 5}{- 2 \cdot 7}$

Cancel the common factor.

$\frac{- 2 \cdot 5}{- 2 \cdot 7}$

Rewrite the expression.

$\frac{5}{7}$

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

$(- 5, \frac{5}{7})$

Find the Holes in the Graph (x^2-25)/(x^2-4x-45)

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