Find the Holes in the Graph (4x^2-100)/(2x^2-7x-15)

$\frac{4 x^{2} - 100}{2 x^{2} - 7 x - 15}$

Step 1

Factor $4 x^{2} - 100$ .

Factor $4$ out of $4 x^{2} - 100$ .

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

$\frac{4 (x^{2}) - 100}{2 x^{2} - 7 x - 15}$

Factor $4$ out of $- 100$ .

$\frac{4 x^{2} + 4 \cdot - 25}{2 x^{2} - 7 x - 15}$

Factor $4$ out of $4 x^{2} + 4 \cdot - 25$ .

$\frac{4 (x^{2} - 25)}{2 x^{2} - 7 x - 15}$

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

$\frac{4 (x^{2} - 5^{2})}{2 x^{2} - 7 x - 15}$

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 $b = 5$ .

$\frac{4 ((x + 5) (x - 5))}{2 x^{2} - 7 x - 15}$

Remove unnecessary parentheses.

$\frac{4 (x + 5) (x - 5)}{2 x^{2} - 7 x - 15}$

Step 2

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 15 = - 30$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$\frac{4 (x + 5) (x - 5)}{2 x^{2} - 7 (x) - 15}$

Rewrite $- 7$ as $3$ plus $- 10$

$\frac{4 (x + 5) (x - 5)}{2 x^{2} + (3 - 10) x - 15}$

Apply the distributive property.

$\frac{4 (x + 5) (x - 5)}{2 x^{2} + 3 x - 10 x - 15}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{4 (x + 5) (x - 5)}{(2 x^{2} + 3 x) - 10 x - 15}$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $2 x + 3$ .

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

Step 3

Cancel the common factor of $x - 5$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Substitute $5$ for $x$ in $\frac{4 (x + 5)}{2 x + 3}$ and simplify.

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

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

Simplify.

Add $5$ and $5$ .

$\frac{4 \cdot 10}{2 \cdot 5 + 3}$

Simplify the denominator.

Multiply $2$ by $5$ .

$\frac{4 \cdot 10}{10 + 3}$

Add $10$ and $3$ .

$\frac{4 \cdot 10}{13}$

Multiply $4$ by $10$ .

$\frac{40}{13}$

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

$(5, \frac{40}{13})$

Step 6

Find the Holes in the Graph (4x^2-100)/(2x^2-7x-15)

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