Find the Holes in the Graph y=(3x^2-12)/(8x^2-18x+4)

$y = \frac{3 x^{2} - 12}{8 x^{2} - 18 x + 4}$

Factor $3 x^{2} - 12$ .

Factor $3$ out of $3 x^{2} - 12$ .

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

$y = \frac{3 (x^{2}) - 12}{8 x^{2} - 18 x + 4}$

Factor $3$ out of $- 12$ .

$y = \frac{3 x^{2} + 3 \cdot - 4}{8 x^{2} - 18 x + 4}$

Factor $3$ out of $3 x^{2} + 3 \cdot - 4$ .

$y = \frac{3 (x^{2} - 4)}{8 x^{2} - 18 x + 4}$

Rewrite $4$ as $2^{2}$ .

$y = \frac{3 (x^{2} - 2^{2})}{8 x^{2} - 18 x + 4}$

Factor.

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 $b = 2$ .

$y = \frac{3 ((x + 2) (x - 2))}{8 x^{2} - 18 x + 4}$

Remove unnecessary parentheses.

$y = \frac{3 (x + 2) (x - 2)}{8 x^{2} - 18 x + 4}$

Factor $8 x^{2} - 18 x + 4$ .

Factor $2$ out of $8 x^{2} - 18 x + 4$ .

Factor $2$ out of $8 x^{2}$ .

$y = \frac{3 (x + 2) (x - 2)}{2 (4 x^{2}) - 18 x + 4}$

Factor $2$ out of $- 18 x$ .

$y = \frac{3 (x + 2) (x - 2)}{2 (4 x^{2}) + 2 (- 9 x) + 4}$

Factor $2$ out of $4$ .

$y = \frac{3 (x + 2) (x - 2)}{2 (4 x^{2}) + 2 (- 9 x) + 2 (2)}$

Factor $2$ out of $2 (4 x^{2}) + 2 (- 9 x)$ .

$y = \frac{3 (x + 2) (x - 2)}{2 (4 x^{2} - 9 x) + 2 (2)}$

Factor $2$ out of $2 (4 x^{2} - 9 x) + 2 (2)$ .

$y = \frac{3 (x + 2) (x - 2)}{2 (4 x^{2} - 9 x + 2)}$

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 $a \cdot c = 4 \cdot 2 = 8$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 x$ .

$y = \frac{3 (x + 2) (x - 2)}{2 (4 x^{2} - 9 x + 2)}$

Rewrite $- 9$ as $- 1$ plus $- 8$

$y = \frac{3 (x + 2) (x - 2)}{2 (4 x^{2} + (- 1 - 8) x + 2)}$

Apply the distributive property.

$y = \frac{3 (x + 2) (x - 2)}{2 (4 x^{2} - 1 x - 8 x + 2)}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$y = \frac{3 (x + 2) (x - 2)}{2 ((4 x^{2} - 1 x) - 8 x + 2)}$

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

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

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

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

Remove unnecessary parentheses.

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

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x - 2$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{3 (x + 2)}{2 (4 x - 1)}$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

Substitute $2$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Cancel the common factor of $2 + 2$ and $2$ .

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

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Add $1$ and $1$ .

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

Simplify the denominator.

Multiply $4$ by $2$ .

$\frac{3 \cdot 2}{8 - 1}$

Subtract $1$ from $8$ .

$\frac{3 \cdot 2}{7}$

Multiply $3$ by $2$ .

$\frac{6}{7}$

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

$(2, \frac{6}{7})$

Find the Holes in the Graph y=(3x^2-12)/(8x^2-18x+4)

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