Find the Holes in the Graph f(x)=(x^2-25)/(2x^2+13x+15)

$f (x) = \frac{x^{2} - 25}{2 x^{2} + 13 x + 15}$

Step 1

Factor $x^{2} - 25$ .

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

$f (x) = \frac{x^{2} - 5^{2}}{2 x^{2} + 13 x + 15}$

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 $b = 5$ .

$f (x) = \frac{(x + 5) (x - 5)}{2 x^{2} + 13 x + 15}$

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 $b = 13$ .

Factor $13$ out of $13 x$ .

$f (x) = \frac{(x + 5) (x - 5)}{2 x^{2} + 13 (x) + 15}$

Rewrite $13$ as $3$ plus $10$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Step 3

Cancel the common factor of $x + 5$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{x - 5}{2 x + 3}$ .

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

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

$\frac{- 5 - 5}{2 \cdot - 5 + 3}$

Simplify.

Subtract $5$ from $- 5$ .

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

Simplify the denominator.

Multiply $2$ by $- 5$ .

$\frac{- 10}{- 10 + 3}$

Add $- 10$ and $3$ .

$\frac{- 10}{- 7}$

Dividing two negative values results in a positive value.

$\frac{10}{7}$

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

$(- 5, \frac{10}{7})$

Step 6

Find the Holes in the Graph f(x)=(x^2-25)/(2x^2+13x+15)

Related Questions

Support LCMGCF.com by disabling your adblocker.