Find the Holes in the Graph f(x)=(x^2+7x+10)/(2x^2-x-10)

$f (x) = \frac{x^{2} + 7 x + 10}{2 x^{2} - x - 10}$

Step 1

Factor $x^{2} + 7 x + 10$ 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 $10$ and whose sum is $7$ .

$2, 5$

Write the factored form using these integers.

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

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 $a \cdot c = 2 \cdot - 10 = - 20$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

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

Rewrite $- 1$ as $4$ plus $- 5$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Step 3

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

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

Simplify.

Add $- 2$ and $5$ .

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

Simplify the denominator.

Multiply $2$ by $- 2$ .

$\frac{3}{- 4 - 5}$

Subtract $5$ from $- 4$ .

$\frac{3}{- 9}$

Cancel the common factor of $3$ and $- 9$ .

Factor $3$ out of $3$ .

$\frac{3 (1)}{- 9}$

Cancel the common factors.

Factor $3$ out of $- 9$ .

$\frac{3 \cdot 1}{3 \cdot - 3}$

Cancel the common factor.

$\frac{3 \cdot 1}{3 \cdot - 3}$

Rewrite the expression.

$\frac{1}{- 3}$

Move the negative in front of the fraction.

$- \frac{1}{3}$

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

$(- 2, - \frac{1}{3})$

Step 6

Find the Holes in the Graph f(x)=(x^2+7x+10)/(2x^2-x-10)

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