Find the Holes in the Graph f(x)=(3x^2-48x+180)/(x^2-4x-12)

$f (x) = \frac{3 x^{2} - 48 x + 180}{x^{2} - 4 x - 12}$

Factor $3 x^{2} - 48 x + 180$ .

Factor $3$ out of $3 x^{2} - 48 x + 180$ .

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

$f (x) = \frac{3 (x^{2}) - 48 x + 180}{x^{2} - 4 x - 12}$

Factor $3$ out of $- 48 x$ .

$f (x) = \frac{3 (x^{2}) + 3 (- 16 x) + 180}{x^{2} - 4 x - 12}$

Factor $3$ out of $180$ .

$f (x) = \frac{3 x^{2} + 3 (- 16 x) + 3 \cdot 60}{x^{2} - 4 x - 12}$

Factor $3$ out of $3 x^{2} + 3 (- 16 x)$ .

$f (x) = \frac{3 (x^{2} - 16 x) + 3 \cdot 60}{x^{2} - 4 x - 12}$

Factor $3$ out of $3 (x^{2} - 16 x) + 3 \cdot 60$ .

$f (x) = \frac{3 (x^{2} - 16 x + 60)}{x^{2} - 4 x - 12}$

Factor.

Factor $x^{2} - 16 x + 60$ 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 $60$ and whose sum is $- 16$ .

$- 10, - 6$

Write the factored form using these integers.

$f (x) = \frac{3 ((x - 10) (x - 6))}{x^{2} - 4 x - 12}$

Remove unnecessary parentheses.

$f (x) = \frac{3 (x - 10) (x - 6)}{x^{2} - 4 x - 12}$

Factor $x^{2} - 4 x - 12$ 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 $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Write the factored form using these integers.

$f (x) = \frac{3 (x - 10) (x - 6)}{(x - 6) (x + 2)}$

Cancel the common factor of $x - 6$ .

Cancel the common factor.

$f (x) = \frac{3 (x - 10) (x - 6)}{(x - 6) (x + 2)}$

Rewrite the expression.

$f (x) = \frac{3 (x - 10)}{x + 2}$

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

$x - 6$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{3 (x - 10)}{x + 2}$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Substitute $6$ for $x$ in $\frac{3 (x - 10)}{x + 2}$ and simplify.

Substitute $6$ for $x$ to find the $y$ coordinate of the hole.

$\frac{3 (6 - 10)}{6 + 2}$

Simplify.

Cancel the common factor of $6 - 10$ and $6 + 2$ .

Factor $2$ out of $3 (6 - 10)$ .

$\frac{2 (3 (3 - 5))}{6 + 2}$

Cancel the common factors.

Factor $2$ out of $6$ .

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 \cdot 3 + 2 (1)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Subtract $5$ from $3$ .

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

Add $3$ and $1$ .

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

Multiply $3$ by $- 2$ .

$\frac{- 6}{4}$

Cancel the common factor of $- 6$ and $4$ .

Factor $2$ out of $- 6$ .

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

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{- 3}{2}$

Move the negative in front of the fraction.

$- \frac{3}{2}$

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

$(6, - \frac{3}{2})$

Find the Holes in the Graph f(x)=(3x^2-48x+180)/(x^2-4x-12)

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