Find the Holes in the Graph f(x)=(5x^2-5)/(-3x^2-6x+9)

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

Factor $5 x^{2} - 5$ .

Factor $5$ out of $5 x^{2} - 5$ .

Factor $5$ out of $5 x^{2}$ .

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

Factor $5$ out of $- 5$ .

$f (x) = \frac{5 (x^{2}) + 5 (- 1)}{- 3 x^{2} - 6 x + 9}$

Factor $5$ out of $5 (x^{2}) + 5 (- 1)$ .

$f (x) = \frac{5 (x^{2} - 1)}{- 3 x^{2} - 6 x + 9}$

Rewrite $1$ as $1^{2}$ .

$f (x) = \frac{5 (x^{2} - 1^{2})}{- 3 x^{2} - 6 x + 9}$

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 $b = 1$ .

$f (x) = \frac{5 ((x + 1) (x - 1))}{- 3 x^{2} - 6 x + 9}$

Remove unnecessary parentheses.

$f (x) = \frac{5 (x + 1) (x - 1)}{- 3 x^{2} - 6 x + 9}$

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

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

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

$f (x) = \frac{5 (x + 1) (x - 1)}{3 (- x^{2}) - 6 x + 9}$

Factor $3$ out of $- 6 x$ .

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

Factor $3$ out of $9$ .

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

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

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

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

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

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

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

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

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

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Remove unnecessary parentheses.

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

Cancel the common factor of $x - 1$ and $- x + 1$ .

Factor $- 1$ out of $x$ .

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

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

$f (x) = \frac{5 (x + 1) (- 1 (- x) - 1 \cdot 1)}{3 (- x + 1) (x + 3)}$

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

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

Cancel the common factor.

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

Rewrite the expression.

$f (x) = \frac{5 (x + 1) \cdot (- 1)}{3 (x + 3)}$

Multiply $- 1$ by $5$ .

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

Move the negative in front of the fraction.

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

Multiply $5$ by $x + 1$ .

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

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

$- x + 1$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{5 (x + 1)}{3 (x + 3)}$ .

Set $- x + 1$ equal to $0$ .

$- x + 1 = 0$

Solve $- x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

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

Substitute $1$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Add $1$ and $1$ .

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

Add $1$ and $3$ .

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

Multiply $5$ by $2$ .

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

Multiply $3$ by $4$ .

$- \frac{10}{12}$

Cancel the common factor of $10$ and $12$ .

Factor $2$ out of $10$ .

$- \frac{2 (5)}{12}$

Cancel the common factors.

Factor $2$ out of $12$ .

$- \frac{2 \cdot 5}{2 \cdot 6}$

Cancel the common factor.

$- \frac{2 \cdot 5}{2 \cdot 6}$

Rewrite the expression.

$- \frac{5}{6}$

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

$(1, - \frac{5}{6})$

Find the Holes in the Graph f(x)=(5x^2-5)/(-3x^2-6x+9)

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