Find the Holes in the Graph (x^2+x-6)/(-4x^2-16x-12)

$\frac{x^{2} + x - 6}{- 4 x^{2} - 16 x - 12}$

Step 1

Factor $x^{2} + x - 6$ 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 $1$ .

$- 2, 3$

Write the factored form using these integers.

$\frac{(x - 2) (x + 3)}{- 4 x^{2} - 16 x - 12}$

Step 2

Factor $- 4 x^{2} - 16 x - 12$ .

Factor $4$ out of $- 4 x^{2} - 16 x - 12$ .

Factor $4$ out of $- 4 x^{2}$ .

$\frac{(x - 2) (x + 3)}{4 (- x^{2}) - 16 x - 12}$

Factor $4$ out of $- 16 x$ .

$\frac{(x - 2) (x + 3)}{4 (- x^{2}) + 4 (- 4 x) - 12}$

Factor $4$ out of $- 12$ .

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

Factor $4$ out of $4 (- x^{2}) + 4 (- 4 x)$ .

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

Factor $4$ out of $4 (- x^{2} - 4 x) + 4 (- 3)$ .

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

Factor.

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 $b = - 4$ .

Factor $- 4$ out of $- 4 x$ .

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

Rewrite $- 4$ as $- 1$ plus $- 3$

$\frac{(x - 2) (x + 3)}{4 (- x^{2} + (- 1 - 3) x - 3)}$

Apply the distributive property.

$\frac{(x - 2) (x + 3)}{4 (- x^{2} - 1 x - 3 x - 3)}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(x - 2) (x + 3)}{4 ((- x^{2} - 1 x) - 3 x - 3)}$

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

$\frac{(x - 2) (x + 3)}{4 (x (- x - 1) + 3 (- x - 1))}$

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

$\frac{(x - 2) (x + 3)}{4 ((- x - 1) (x + 3))}$

Remove unnecessary parentheses.

$\frac{(x - 2) (x + 3)}{4 (- x - 1) (x + 3)}$

Step 3

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$\frac{(x - 2) (x + 3)}{4 (- x - 1) (x + 3)}$

Rewrite the expression.

$\frac{x - 2}{4 (- x - 1)}$

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 $\frac{x - 2}{4 (- x - 1)}$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Substitute $- 3$ for $x$ in $\frac{x - 2}{4 (- x - 1)}$ and simplify.

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

$\frac{- 3 - 2}{4 (- - 3 - 1)}$

Simplify.

Subtract $2$ from $- 3$ .

$\frac{- 5}{4 (- - 3 - 1)}$

Simplify the denominator.

Multiply $- 1$ by $- 3$ .

$\frac{- 5}{4 (3 - 1)}$

Subtract $1$ from $3$ .

$\frac{- 5}{4 \cdot 2}$

Multiply $4$ by $2$ .

$\frac{- 5}{8}$

Move the negative in front of the fraction.

$- \frac{5}{8}$

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

$(- 3, - \frac{5}{8})$

Step 6

Find the Holes in the Graph (x^2+x-6)/(-4x^2-16x-12)

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