Algebra Examples

$\frac{x^{2} - 25}{2 x + 10}$

Step 1

Factor $x^{2} - 25$ .

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Step 1.1

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

$\frac{x^{2} - 5^{2}}{2 x + 10}$

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 $b = 5$ .

$\frac{(x + 5) (x - 5)}{2 x + 10}$

Step 2

Factor $2$ out of $2 x + 10$ .

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Step 2.1

Factor $2$ out of $2 x$ .

$\frac{(x + 5) (x - 5)}{2 (x) + 10}$

Step 2.2

Factor $2$ out of $10$ .

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

Step 2.3

Factor $2$ out of $2 x + 2 \cdot 5$ .

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

Step 3

Cancel the common factor of $x + 5$ .

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Step 3.1

Cancel the common factor.

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

Step 3.2

Rewrite the expression.

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

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

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Step 5.1

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

$x + 5 = 0$

Step 5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 5.3

Substitute $- 5$ for $x$ in $\frac{x - 5}{2}$ and simplify.

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Step 5.3.1

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

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

Step 5.3.2

Simplify.

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Step 5.3.2.1

Subtract $5$ from $- 5$ .

$\frac{- 10}{2}$

Step 5.3.2.2

Divide $- 10$ by $2$ .

$- 5$

Step 5.4

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

$(- 5, - 5)$

Step 6