Find the Holes in the Graph f(x)=(-x^2-2x+3)/(x^2+5x+6)

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

Step 1

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 $a \cdot c = - 1 \cdot 3 = - 3$ and whose sum is $b = - 2$ .

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

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

Rewrite $- 2$ as $1$ plus $- 3$

$f (x) = \frac{- x^{2} + (1 - 3) x + 3}{x^{2} + 5 x + 6}$

Apply the distributive property.

$f (x) = \frac{- x^{2} + 1 x - 3 x + 3}{x^{2} + 5 x + 6}$

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$f (x) = \frac{x (- x + 1) + 3 (- x + 1)}{x^{2} + 5 x + 6}$

Factor the polynomial by factoring out the greatest common factor, $- x + 1$ .

$f (x) = \frac{(- x + 1) (x + 3)}{x^{2} + 5 x + 6}$

Step 2

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.

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

Step 3

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{- x + 1}{x + 2}$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

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

Simplify.

Add $3$ and $1$ .

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

Add $- 3$ and $2$ .

$\frac{4}{- 1}$

Divide $4$ by $- 1$ .

$- 4$

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

$(- 3, - 4)$

Step 6

Find the Holes in the Graph f(x)=(-x^2-2x+3)/(x^2+5x+6)

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