Find the Holes in the Graph f(x)=(2x^2+10x+12)/(x^2+3x+2)

$f (x) = \frac{2 x^{2} + 10 x + 12}{x^{2} + 3 x + 2}$

Step 1

Factor $2 x^{2} + 10 x + 12$ .

Factor $2$ out of $2 x^{2} + 10 x + 12$ .

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

$f (x) = \frac{2 (x^{2}) + 10 x + 12}{x^{2} + 3 x + 2}$

Factor $2$ out of $10 x$ .

$f (x) = \frac{2 (x^{2}) + 2 (5 x) + 12}{x^{2} + 3 x + 2}$

Factor $2$ out of $12$ .

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

Factor $2$ out of $2 x^{2} + 2 (5 x)$ .

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

Factor $2$ out of $2 (x^{2} + 5 x) + 2 \cdot 6$ .

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

Factor.

Factor $x^{2} + 5 x + 6$ 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 $6$ and whose sum is $5$ .

$2, 3$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 2

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

$1, 2$

Write the factored form using these integers.

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

Step 3

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

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

Simplify.

Add $- 2$ and $3$ .

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

Add $- 2$ and $1$ .

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

Multiply $2$ by $1$ .

$\frac{2}{- 1}$

Divide $2$ by $- 1$ .

$- 2$

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

$(- 2, - 2)$

Step 6

Find the Holes in the Graph f(x)=(2x^2+10x+12)/(x^2+3x+2)

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