Find the Holes in the Graph f(x)=(2x^2-2x-24)/(x^2+3x)

$f (x) = \frac{2 x^{2} - 2 x - 24}{x^{2} + 3 x}$

Factor $2 x^{2} - 2 x - 24$ .

Factor $2$ out of $2 x^{2} - 2 x - 24$ .

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

$f (x) = \frac{2 (x^{2}) - 2 x - 24}{x^{2} + 3 x}$

Factor $2$ out of $- 2 x$ .

$f (x) = \frac{2 (x^{2}) + 2 (- x) - 24}{x^{2} + 3 x}$

Factor $2$ out of $- 24$ .

$f (x) = \frac{2 (x^{2}) + 2 (- x) + 2 (- 12)}{x^{2} + 3 x}$

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

$f (x) = \frac{2 (x^{2} - x) + 2 (- 12)}{x^{2} + 3 x}$

Factor $2$ out of $2 (x^{2} - x) + 2 (- 12)$ .

$f (x) = \frac{2 (x^{2} - x - 12)}{x^{2} + 3 x}$

Factor.

Factor $x^{2} - x - 12$ 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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

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

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

Factor $x$ out of $3 x$ .

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

Factor $x$ out of $x \cdot x + x \cdot 3$ .

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

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x + 3$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

$\frac{2 (- 3 - 4)}{- 3}$

Simplify.

Subtract $4$ from $- 3$ .

$\frac{2 \cdot - 7}{- 3}$

Multiply $2$ by $- 7$ .

$\frac{- 14}{- 3}$

Dividing two negative values results in a positive value.

$\frac{14}{3}$

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

$(- 3, \frac{14}{3})$

Find the Holes in the Graph f(x)=(2x^2-2x-24)/(x^2+3x)

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