Find the Holes in the Graph y=(x^2-9)/(2x^2+3x-27)

$y = \frac{x^{2} - 9}{2 x^{2} + 3 x - 27}$

Step 1

Factor $x^{2} - 9$ .

Rewrite $9$ as $3^{2}$ .

$y = \frac{x^{2} - 3^{2}}{2 x^{2} + 3 x - 27}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$y = \frac{(x + 3) (x - 3)}{2 x^{2} + 3 x - 27}$

Step 2

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 $a \cdot c = 2 \cdot - 27 = - 54$ and whose sum is $b = 3$ .

Factor $3$ out of $3 x$ .

$y = \frac{(x + 3) (x - 3)}{2 x^{2} + 3 (x) - 27}$

Rewrite $3$ as $- 6$ plus $9$

$y = \frac{(x + 3) (x - 3)}{2 x^{2} + (- 6 + 9) x - 27}$

Apply the distributive property.

$y = \frac{(x + 3) (x - 3)}{2 x^{2} - 6 x + 9 x - 27}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$y = \frac{(x + 3) (x - 3)}{(2 x^{2} - 6 x) + 9 x - 27}$

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

$y = \frac{(x + 3) (x - 3)}{2 x (x - 3) + 9 (x - 3)}$

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

$y = \frac{(x + 3) (x - 3)}{(x - 3) (2 x + 9)}$

Step 3

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$y = \frac{(x + 3) (x - 3)}{(x - 3) (2 x + 9)}$

Rewrite the expression.

$y = \frac{x + 3}{2 x + 9}$

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 $\frac{x + 3}{2 x + 9}$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Substitute $3$ for $x$ in $\frac{x + 3}{2 x + 9}$ and simplify.

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

$\frac{3 + 3}{2 \cdot 3 + 9}$

Simplify.

Cancel the common factor of $3 + 3$ and $2 \cdot 3 + 9$ .

Factor $3$ out of $3$ .

$\frac{3 \cdot 1 + 3}{2 \cdot 3 + 9}$

Factor $3$ out of $3$ .

$\frac{3 \cdot 1 + 3 \cdot 1}{2 \cdot 3 + 9}$

Factor $3$ out of $3 \cdot 1 + 3 \cdot 1$ .

$\frac{3 \cdot (1 + 1)}{2 \cdot 3 + 9}$

Cancel the common factors.

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

$\frac{3 \cdot (1 + 1)}{3 \cdot 2 + 9}$

Factor $3$ out of $9$ .

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Add $1$ and $1$ .

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

Add $2$ and $3$ .

$\frac{2}{5}$

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

$(3, \frac{2}{5})$

Step 6

Find the Holes in the Graph y=(x^2-9)/(2x^2+3x-27)

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