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

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

Step 1

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

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $6$ .

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

Factor $2$ out of $2 x + 2 \cdot 3$ .

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

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.

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

Step 3

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x + 3$

Step 5

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

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

Simplify.

Add $- 3$ and $1$ .

$\frac{2}{- 2}$

Divide $2$ by $- 2$ .

$- 1$

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

$(- 3, - 1)$

Step 6

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

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