Find the Holes in the Graph (4x^2-16)/(5x^2+15x+10)

$\frac{4 x^{2} - 16}{5 x^{2} + 15 x + 10}$

Step 1

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

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

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

$\frac{4 (x^{2}) - 16}{5 x^{2} + 15 x + 10}$

Factor $4$ out of $- 16$ .

$\frac{4 x^{2} + 4 \cdot - 4}{5 x^{2} + 15 x + 10}$

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

$\frac{4 (x^{2} - 4)}{5 x^{2} + 15 x + 10}$

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

$\frac{4 (x^{2} - 2^{2})}{5 x^{2} + 15 x + 10}$

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 $b = 2$ .

$\frac{4 ((x + 2) (x - 2))}{5 x^{2} + 15 x + 10}$

Remove unnecessary parentheses.

$\frac{4 (x + 2) (x - 2)}{5 x^{2} + 15 x + 10}$

Step 2

Factor $5 x^{2} + 15 x + 10$ .

Factor $5$ out of $5 x^{2} + 15 x + 10$ .

Factor $5$ out of $5 x^{2}$ .

$\frac{4 (x + 2) (x - 2)}{5 (x^{2}) + 15 x + 10}$

Factor $5$ out of $15 x$ .

$\frac{4 (x + 2) (x - 2)}{5 (x^{2}) + 5 (3 x) + 10}$

Factor $5$ out of $10$ .

$\frac{4 (x + 2) (x - 2)}{5 x^{2} + 5 (3 x) + 5 \cdot 2}$

Factor $5$ out of $5 x^{2} + 5 (3 x)$ .

$\frac{4 (x + 2) (x - 2)}{5 (x^{2} + 3 x) + 5 \cdot 2}$

Factor $5$ out of $5 (x^{2} + 3 x) + 5 \cdot 2$ .

$\frac{4 (x + 2) (x - 2)}{5 (x^{2} + 3 x + 2)}$

Factor.

Factor $x^{2} + 3 x + 2$ using the AC method.

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 $2$ and whose sum is $3$ .

$1, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 3

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x + 2$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{4 (x - 2)}{5 (x + 1)}$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Substitute $- 2$ for $x$ in $\frac{4 (x - 2)}{5 (x + 1)}$ and simplify.

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

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

Simplify.

Subtract $2$ from $- 2$ .

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

Add $- 2$ and $1$ .

$\frac{4 \cdot - 4}{5 \cdot - 1}$

Multiply $4$ by $- 4$ .

$\frac{- 16}{5 \cdot - 1}$

Multiply $5$ by $- 1$ .

$\frac{- 16}{- 5}$

Dividing two negative values results in a positive value.

$\frac{16}{5}$

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

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

Step 6

Find the Holes in the Graph (4x^2-16)/(5x^2+15x+10)

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