Find the Holes in the Graph (-3x^2-18x-15)/(2x^2+9x-5)

$\frac{- 3 x^{2} - 18 x - 15}{2 x^{2} + 9 x - 5}$

Factor $- 3 x^{2} - 18 x - 15$ .

Factor $3$ out of $- 3 x^{2} - 18 x - 15$ .

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

$\frac{3 (- x^{2}) - 18 x - 15}{2 x^{2} + 9 x - 5}$

Factor $3$ out of $- 18 x$ .

$\frac{3 (- x^{2}) + 3 (- 6 x) - 15}{2 x^{2} + 9 x - 5}$

Factor $3$ out of $- 15$ .

$\frac{3 (- x^{2}) + 3 (- 6 x) + 3 (- 5)}{2 x^{2} + 9 x - 5}$

Factor $3$ out of $3 (- x^{2}) + 3 (- 6 x)$ .

$\frac{3 (- x^{2} - 6 x) + 3 (- 5)}{2 x^{2} + 9 x - 5}$

Factor $3$ out of $3 (- x^{2} - 6 x) + 3 (- 5)$ .

$\frac{3 (- x^{2} - 6 x - 5)}{2 x^{2} + 9 x - 5}$

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 $a \cdot c = - 1 \cdot - 5 = 5$ and whose sum is $b = - 6$ .

Factor $- 6$ out of $- 6 x$ .

$\frac{3 (- x^{2} - 6 (x) - 5)}{2 x^{2} + 9 x - 5}$

Rewrite $- 6$ as $- 1$ plus $- 5$

$\frac{3 (- x^{2} + (- 1 - 5) x - 5)}{2 x^{2} + 9 x - 5}$

Apply the distributive property.

$\frac{3 (- x^{2} - 1 x - 5 x - 5)}{2 x^{2} + 9 x - 5}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{3 ((- x^{2} - 1 x) - 5 x - 5)}{2 x^{2} + 9 x - 5}$

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

$\frac{3 (x (- x - 1) + 5 (- x - 1))}{2 x^{2} + 9 x - 5}$

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

$\frac{3 ((- x - 1) (x + 5))}{2 x^{2} + 9 x - 5}$

Remove unnecessary parentheses.

$\frac{3 (- x - 1) (x + 5)}{2 x^{2} + 9 x - 5}$

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 $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = 9$ .

Factor $9$ out of $9 x$ .

$\frac{3 (- x - 1) (x + 5)}{2 x^{2} + 9 (x) - 5}$

Rewrite $9$ as $- 1$ plus $10$

$\frac{3 (- x - 1) (x + 5)}{2 x^{2} + (- 1 + 10) x - 5}$

Apply the distributive property.

$\frac{3 (- x - 1) (x + 5)}{2 x^{2} - 1 x + 10 x - 5}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{3 (- x - 1) (x + 5)}{(2 x^{2} - 1 x) + 10 x - 5}$

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

$\frac{3 (- x - 1) (x + 5)}{x (2 x - 1) + 5 (2 x - 1)}$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$\frac{3 (- x - 1) (x + 5)}{(2 x - 1) (x + 5)}$

Cancel the common factor of $x + 5$ .

Cancel the common factor.

$\frac{3 (- x - 1) (x + 5)}{(2 x - 1) (x + 5)}$

Rewrite the expression.

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

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

$x + 5$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{3 (- x - 1)}{2 x - 1}$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Substitute $- 5$ for $x$ in $\frac{3 (- x - 1)}{2 x - 1}$ and simplify.

Substitute $- 5$ for $x$ to find the $y$ coordinate of the hole.

$\frac{3 (5 - 1)}{2 \cdot - 5 - 1}$

Simplify.

Subtract $1$ from $5$ .

$\frac{3 \cdot 4}{2 \cdot - 5 - 1}$

Simplify the denominator.

Multiply $2$ by $- 5$ .

$\frac{3 \cdot 4}{- 10 - 1}$

Subtract $1$ from $- 10$ .

$\frac{3 \cdot 4}{- 11}$

Multiply $3$ by $4$ .

$\frac{12}{- 11}$

Move the negative in front of the fraction.

$- \frac{12}{11}$

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

$(- 5, - \frac{12}{11})$

Find the Holes in the Graph (-3x^2-18x-15)/(2x^2+9x-5)

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