Find the Holes in the Graph (x-5)/(x^2-25)

$\frac{x - 5}{x^{2} - 25}$

Step 1

Factor $x^{2} - 25$ .

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

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

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 $b = 5$ .

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

Step 2

Cancel the common factor of $x - 5$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{x + 5}$

Step 3

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

$x - 5$

Step 4

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{1}{x + 5}$ .

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Substitute $5$ for $x$ in $\frac{1}{x + 5}$ and simplify.

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

$\frac{1}{5 + 5}$

Add $5$ and $5$ .

$\frac{1}{10}$

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

$(5, \frac{1}{10})$

Step 5