Find the Holes in the Graph (4x^2-16)/(-5x+10)

$\frac{4 x^{2} - 16}{- 5 x + 10}$

Step 1

Factor $4 x^{2} - 16$ .

Factor $4$ out of $4 x^{2} - 16$ .

Factor $4$ out of $4 x^{2}$ .

$\frac{4 (x^{2}) - 16}{- 5 x + 10}$

Factor $4$ out of $- 16$ .

$\frac{4 x^{2} + 4 \cdot - 4}{- 5 x + 10}$

Factor $4$ out of $4 x^{2} + 4 \cdot - 4$ .

$\frac{4 (x^{2} - 4)}{- 5 x + 10}$

Rewrite $4$ as $2^{2}$ .

$\frac{4 (x^{2} - 2^{2})}{- 5 x + 10}$

Factor.

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 $b = 2$ .

$\frac{4 ((x + 2) (x - 2))}{- 5 x + 10}$

Remove unnecessary parentheses.

$\frac{4 (x + 2) (x - 2)}{- 5 x + 10}$

Step 2

Factor $5$ out of $- 5 x + 10$ .

Factor $5$ out of $- 5 x$ .

$\frac{4 (x + 2) (x - 2)}{5 (- x) + 10}$

Factor $5$ out of $10$ .

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

Factor $5$ out of $5 (- x) + 5 (2)$ .

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

Step 3

Cancel the common factor of $x - 2$ and $- x + 2$ .

Factor $- 1$ out of $x$ .

$\frac{4 (x + 2) (- 1 (- x) - 2)}{5 (- x + 2)}$

Rewrite $- 2$ as $- 1 (2)$ .

$\frac{4 (x + 2) (- 1 (- x) - 1 (2))}{5 (- x + 2)}$

Factor $- 1$ out of $- 1 (- x) - 1 (2)$ .

$\frac{4 (x + 2) (- 1 (- x + 2))}{5 (- x + 2)}$

Cancel the common factor.

$\frac{4 (x + 2) (- 1 (- x + 2))}{5 (- x + 2)}$

Rewrite the expression.

$\frac{4 (x + 2) \cdot (- 1)}{5}$

Step 4

Multiply $- 1$ by $4$ .

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

Step 5

Move the negative in front of the fraction.

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

Step 6

Multiply $4$ by $x + 2$ .

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

Step 7

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

$- x + 2$

Step 8

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{4 (x + 2)}{5}$ .

Set $- x + 2$ equal to $0$ .

$- x + 2 = 0$

Solve $- x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$- x = - 2$

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x = 2$

Substitute $2$ for $x$ in $- \frac{4 (x + 2)}{5}$ and simplify.

Substitute $2$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Add $2$ and $2$ .

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

Multiply $4$ by $4$ .

$- \frac{16}{5}$

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

$(2, - \frac{16}{5})$

Step 9

Find the Holes in the Graph (4x^2-16)/(-5x+10)

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