Find the Holes in the Graph (x^2-25)/(4x+20)

$\frac{x^{2} - 25}{4 x + 20}$

Step 1

Factor $x^{2} - 25$ .

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

$\frac{x^{2} - 5^{2}}{4 x + 20}$

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 $b = 5$ .

$\frac{(x + 5) (x - 5)}{4 x + 20}$

Step 2

Factor $4$ out of $4 x + 20$ .

Factor $4$ out of $4 x$ .

$\frac{(x + 5) (x - 5)}{4 (x) + 20}$

Factor $4$ out of $20$ .

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

Factor $4$ out of $4 x + 4 \cdot 5$ .

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

Step 3

Cancel the common factor of $x + 5$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{x - 5}{4}$ .

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

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

$\frac{- 5 - 5}{4}$

Simplify.

Subtract $5$ from $- 5$ .

$\frac{- 10}{4}$

Cancel the common factor of $- 10$ and $4$ .

Factor $2$ out of $- 10$ .

$\frac{2 (- 5)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{- 5}{2}$

Move the negative in front of the fraction.

$- \frac{5}{2}$

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

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

Step 6

Find the Holes in the Graph (x^2-25)/(4x+20)

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