Algebra Examples

$\frac{x^{2} - 25}{x^{2} - x - 20}$

Step 1

Factor $x^{2} - 25$ .

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Step 1.1

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

$\frac{x^{2} - 5^{2}}{x^{2} - x - 20}$

Step 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 $b = 5$ .

$\frac{(x + 5) (x - 5)}{x^{2} - x - 20}$

Step 2

Factor $x^{2} - x - 20$ using the AC method.

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Step 2.1

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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Step 2.2

Write the factored form using these integers.

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

Step 3

Cancel the common factor of $x - 5$ .

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Step 3.1

Cancel the common factor.

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

Step 3.2

Rewrite the expression.

$\frac{x + 5}{x + 4}$

Step 4

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

$x - 5$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 5}{x + 4}$ .

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Step 5.1

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

$x - 5 = 0$

Step 5.2

Add $5$ to both sides of the equation.

$x = 5$

Step 5.3

Substitute $5$ for $x$ in $\frac{x + 5}{x + 4}$ and simplify.

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Step 5.3.1

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

$\frac{5 + 5}{5 + 4}$

Step 5.3.2

Simplify.

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Step 5.3.2.1

Add $5$ and $5$ .

$\frac{10}{5 + 4}$

Step 5.3.2.2

Add $5$ and $4$ .

$\frac{10}{9}$

Step 5.4

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

$(5, \frac{10}{9})$

Step 6