Find the Holes in the Graph f(x)=(-3x^2+9x-6)/(2x^2-8)

$f (x) = \frac{- 3 x^{2} + 9 x - 6}{2 x^{2} - 8}$

Step 1

Factor $- 3 x^{2} + 9 x - 6$ .

Factor $3$ out of $- 3 x^{2} + 9 x - 6$ .

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

$f (x) = \frac{3 (- x^{2}) + 9 x - 6}{2 x^{2} - 8}$

Factor $3$ out of $9 x$ .

$f (x) = \frac{3 (- x^{2}) + 3 (3 x) - 6}{2 x^{2} - 8}$

Factor $3$ out of $- 6$ .

$f (x) = \frac{3 (- x^{2}) + 3 (3 x) + 3 (- 2)}{2 x^{2} - 8}$

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

$f (x) = \frac{3 (- x^{2} + 3 x) + 3 (- 2)}{2 x^{2} - 8}$

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

$f (x) = \frac{3 (- x^{2} + 3 x - 2)}{2 x^{2} - 8}$

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 $a \cdot c = - 1 \cdot - 2 = 2$ and whose sum is $b = 3$ .

Factor $3$ out of $3 x$ .

$f (x) = \frac{3 (- x^{2} + 3 (x) - 2)}{2 x^{2} - 8}$

Rewrite $3$ as $1$ plus $2$

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

$f (x) = \frac{3 (- x^{2} + x + 2 x - 2)}{2 x^{2} - 8}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{3 ((- x^{2} + x) + 2 x - 2)}{2 x^{2} - 8}$

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

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

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

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

Remove unnecessary parentheses.

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

Step 2

Factor $2 x^{2} - 8$ .

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

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

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

Factor $2$ out of $- 8$ .

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

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

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

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

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

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 $b = 2$ .

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

Remove unnecessary parentheses.

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

Step 3

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x - 2$

Step 5

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

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

Simplify.

Add $- 2$ and $1$ .

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

Add $2$ and $2$ .

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

Multiply $3$ by $- 1$ .

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

Multiply $2$ by $4$ .

$\frac{- 3}{8}$

Move the negative in front of the fraction.

$- \frac{3}{8}$

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

$(2, - \frac{3}{8})$

Step 6

Find the Holes in the Graph f(x)=(-3x^2+9x-6)/(2x^2-8)

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