Find the Holes in the Graph f(x)=(-2x^2-6x)/(x^2+4x+3)

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

Step 1

Factor $2 x$ out of $- 2 x^{2} - 6 x$ .

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

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

Factor $2 x$ out of $- 6 x$ .

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

Factor $2 x$ out of $2 x (- x) + 2 x (- 3)$ .

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

Step 2

Factor $x^{2} + 4 x + 3$ 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 $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

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

Step 3

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

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

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

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

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

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Multiply $- 1$ by $2$ .

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

Step 5

Move the negative in front of the fraction.

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

Step 6

Multiply $2$ by $x$ .

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

Step 7

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

$x + 3$

Step 8

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{2 x}{x + 1}$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Substitute $- 3$ for $x$ in $- \frac{2 x}{x + 1}$ and simplify.

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

$- \frac{2 \cdot - 3}{- 3 + 1}$

Simplify.

Multiply $2$ by $- 3$ .

$- \frac{- 6}{- 3 + 1}$

Add $- 3$ and $1$ .

$- \frac{- 6}{- 2}$

Divide $- 6$ by $- 2$ .

$- 1 \cdot 3$

Multiply $- 1$ by $3$ .

$- 3$

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

$(- 3, - 3)$

Step 9

Find the Holes in the Graph f(x)=(-2x^2-6x)/(x^2+4x+3)

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