Find the Holes in the Graph f(x)=(-3x^2-21x-36)/(2x^2+6x)

$f (x) = \frac{- 3 x^{2} - 21 x - 36}{2 x^{2} + 6 x}$

Factor $- 3 x^{2} - 21 x - 36$ .

Factor $3$ out of $- 3 x^{2} - 21 x - 36$ .

Factor $3$ out of $- 3 x^{2}$ .

$f (x) = \frac{3 (- x^{2}) - 21 x - 36}{2 x^{2} + 6 x}$

Factor $3$ out of $- 21 x$ .

$f (x) = \frac{3 (- x^{2}) + 3 (- 7 x) - 36}{2 x^{2} + 6 x}$

Factor $3$ out of $- 36$ .

$f (x) = \frac{3 (- x^{2}) + 3 (- 7 x) + 3 (- 12)}{2 x^{2} + 6 x}$

Factor $3$ out of $3 (- x^{2}) + 3 (- 7 x)$ .

$f (x) = \frac{3 (- x^{2} - 7 x) + 3 (- 12)}{2 x^{2} + 6 x}$

Factor $3$ out of $3 (- x^{2} - 7 x) + 3 (- 12)$ .

$f (x) = \frac{3 (- x^{2} - 7 x - 12)}{2 x^{2} + 6 x}$

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 $a \cdot c = - 1 \cdot - 12 = 12$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$f (x) = \frac{3 (- x^{2} - 7 x - 12)}{2 x^{2} + 6 x}$

Rewrite $- 7$ as $- 3$ plus $- 4$

$f (x) = \frac{3 (- x^{2} + (- 3 - 4) x - 12)}{2 x^{2} + 6 x}$

Apply the distributive property.

$f (x) = \frac{3 (- x^{2} - 3 x - 4 x - 12)}{2 x^{2} + 6 x}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{3 ((- x^{2} - 3 x) - 4 x - 12)}{2 x^{2} + 6 x}$

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

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

Factor the polynomial by factoring out the greatest common factor, $- x - 3$ .

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

Remove unnecessary parentheses.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Combine exponents.

Factor out negative.

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

Multiply $3$ by $- 1$ .

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

Move the negative in front of the fraction.

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

Multiply $3$ by $x + 4$ .

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

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

$x + 3$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{3 (x + 4)}{2 x}$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

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

Simplify.

Cancel the common factors.

Factor $3$ out of $2 \cdot - 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

$- \frac{- 3 + 4}{2 \cdot - 1}$

Add $- 3$ and $4$ .

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

Multiply $2$ by $- 1$ .

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

Move the negative in front of the fraction.

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

Multiply $- - \frac{1}{2}$ .

Multiply $- 1$ by $- 1$ .

$1 (\frac{1}{2})$

Multiply $\frac{1}{2}$ by $1$ .

$\frac{1}{2}$

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

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

Find the Holes in the Graph f(x)=(-3x^2-21x-36)/(2x^2+6x)

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