Find the Holes in the Graph f(x)=(-4x+12)/(2x^2-x-15)

$f (x) = \frac{- 4 x + 12}{2 x^{2} - x - 15}$

Step 1

Factor $4$ out of $- 4 x + 12$ .

Factor $4$ out of $- 4 x$ .

$f (x) = \frac{4 (- x) + 12}{2 x^{2} - x - 15}$

Factor $4$ out of $12$ .

$f (x) = \frac{4 (- x) + 4 (3)}{2 x^{2} - x - 15}$

Factor $4$ out of $4 (- x) + 4 (3)$ .

$f (x) = \frac{4 (- x + 3)}{2 x^{2} - x - 15}$

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 $b = - 1$ .

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

$f (x) = \frac{4 (- x + 3)}{2 x^{2} - (x) - 15}$

Rewrite $- 1$ as $5$ plus $- 6$

$f (x) = \frac{4 (- x + 3)}{2 x^{2} + (5 - 6) x - 15}$

Apply the distributive property.

$f (x) = \frac{4 (- x + 3)}{2 x^{2} + 5 x - 6 x - 15}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{4 (- x + 3)}{(2 x^{2} + 5 x) - 6 x - 15}$

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x + 5$ .

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

Step 3

Cancel the common factor of $- x + 3$ and $x - 3$ .

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

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

Rewrite $3$ as $- 1 (- 3)$ .

$f (x) = \frac{4 (- (x) - 1 \cdot - 3)}{(2 x + 5) (x - 3)}$

Factor $- 1$ out of $- (x) - 1 (- 3)$ .

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

Rewrite $- (x - 3)$ as $- 1 (x - 3)$ .

$f (x) = \frac{4 (- 1 (x - 3))}{(2 x + 5) (x - 3)}$

Cancel the common factor.

$f (x) = \frac{4 (- 1 (x - 3))}{(2 x + 5) (x - 3)}$

Rewrite the expression.

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

Step 4

Simplify the numerator.

Multiply $4$ by $- 1$ .

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

Multiply $4$ by $- 1$ .

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

Step 5

Move the negative in front of the fraction.

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

Step 6

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

$x - 3$

Step 7

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

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

Substitute $3$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify the denominator.

Multiply $2$ by $3$ .

$- \frac{4}{6 + 5}$

Add $6$ and $5$ .

$- \frac{4}{11}$

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

$(3, - \frac{4}{11})$

Step 8

Find the Holes in the Graph f(x)=(-4x+12)/(2x^2-x-15)

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