Find the Holes in the Graph (-4x^2+16x-16)/(3x^2-3x-6)

$\frac{- 4 x^{2} + 16 x - 16}{3 x^{2} - 3 x - 6}$

Step 1

Factor $- 4 x^{2} + 16 x - 16$ .

Factor $4$ out of $- 4 x^{2} + 16 x - 16$ .

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

$\frac{4 (- x^{2}) + 16 x - 16}{3 x^{2} - 3 x - 6}$

Factor $4$ out of $16 x$ .

$\frac{4 (- x^{2}) + 4 (4 x) - 16}{3 x^{2} - 3 x - 6}$

Factor $4$ out of $- 16$ .

$\frac{4 (- x^{2}) + 4 (4 x) + 4 (- 4)}{3 x^{2} - 3 x - 6}$

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

$\frac{4 (- x^{2} + 4 x) + 4 (- 4)}{3 x^{2} - 3 x - 6}$

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

$\frac{4 (- x^{2} + 4 x - 4)}{3 x^{2} - 3 x - 6}$

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 $a \cdot c = - 1 \cdot - 4 = 4$ and whose sum is $b = 4$ .

Factor $4$ out of $4 x$ .

$\frac{4 (- x^{2} + 4 (x) - 4)}{3 x^{2} - 3 x - 6}$

Rewrite $4$ as $2$ plus $2$

$\frac{4 (- x^{2} + (2 + 2) x - 4)}{3 x^{2} - 3 x - 6}$

Apply the distributive property.

$\frac{4 (- x^{2} + 2 x + 2 x - 4)}{3 x^{2} - 3 x - 6}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{4 ((- x^{2} + 2 x) + 2 x - 4)}{3 x^{2} - 3 x - 6}$

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

$\frac{4 (x (- x + 2) - 2 (- x + 2))}{3 x^{2} - 3 x - 6}$

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

$\frac{4 ((- x + 2) (x - 2))}{3 x^{2} - 3 x - 6}$

Remove unnecessary parentheses.

$\frac{4 (- x + 2) (x - 2)}{3 x^{2} - 3 x - 6}$

Combine exponents.

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

$\frac{4 (- (x) + 2) (x - 2)}{3 x^{2} - 3 x - 6}$

Rewrite $2$ as $- 1 (- 2)$ .

$\frac{4 (- (x) - 1 (- 2)) (x - 2)}{3 x^{2} - 3 x - 6}$

Factor $- 1$ out of $- (x) - 1 (- 2)$ .

$\frac{4 (- (x - 2)) (x - 2)}{3 x^{2} - 3 x - 6}$

Rewrite $- (x - 2)$ as $- 1 (x - 2)$ .

$\frac{4 (- 1 (x - 2)) (x - 2)}{3 x^{2} - 3 x - 6}$

Remove parentheses.

$\frac{4 \cdot - 1 (x - 2) (x - 2)}{3 x^{2} - 3 x - 6}$

Raise $x - 2$ to the power of $1$ .

$\frac{4 \cdot - 1 ({(x - 2)}^{1} (x - 2))}{3 x^{2} - 3 x - 6}$

Raise $x - 2$ to the power of $1$ .

$\frac{4 \cdot - 1 ({(x - 2)}^{1} {(x - 2)}^{1})}{3 x^{2} - 3 x - 6}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{4 \cdot - 1 {(x - 2)}^{1 + 1}}{3 x^{2} - 3 x - 6}$

Add $1$ and $1$ .

$\frac{4 \cdot - 1 {(x - 2)}^{2}}{3 x^{2} - 3 x - 6}$

Multiply $4$ by $- 1$ .

$\frac{- 4 {(x - 2)}^{2}}{3 x^{2} - 3 x - 6}$

Step 2

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

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

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

$\frac{- 4 {(x - 2)}^{2}}{3 (x^{2}) - 3 x - 6}$

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

$\frac{- 4 {(x - 2)}^{2}}{3 (x^{2}) + 3 (- x) - 6}$

Factor $3$ out of $- 6$ .

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

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

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

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

$\frac{- 4 {(x - 2)}^{2}}{3 (x^{2} - x - 2)}$

Factor.

Factor $x^{2} - x - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 3

Cancel the common factor of ${(x - 2)}^{2}$ and $x - 2$ .

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

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

Cancel the common factors.

Factor $x - 2$ out of $3 (x - 2) (x + 1)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Move the negative in front of the fraction.

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

Step 5

Multiply $4$ by $x - 2$ .

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

Step 6

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

$x - 2$

Step 7

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

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

Simplify.

Subtract $2$ from $2$ .

$- \frac{4 \cdot 0}{3 (2 + 1)}$

Add $2$ and $1$ .

$- \frac{4 \cdot 0}{3 \cdot 3}$

Multiply $4$ by $0$ .

$- \frac{0}{3 \cdot 3}$

Multiply $3$ by $3$ .

$- \frac{0}{9}$

Divide $0$ by $9$ .

$- 0$

Multiply $- 1$ by $0$ .

$0$

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

$(2, 0)$

Step 8

Find the Holes in the Graph (-4x^2+16x-16)/(3x^2-3x-6)

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