Find the Holes in the Graph (3x-12)/(x^2-16)

$\frac{3 x - 12}{x^{2} - 16}$

Step 1

Factor $3$ out of $3 x - 12$ .

Factor $3$ out of $3 x$ .

$\frac{3 (x) - 12}{x^{2} - 16}$

Factor $3$ out of $- 12$ .

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

Factor $3$ out of $3 x + 3 \cdot - 4$ .

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

Step 2

Factor $x^{2} - 16$ .

Rewrite $16$ as $4^{2}$ .

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

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 $b = 4$ .

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

Step 3

Cancel the common factor of $x - 4$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{3}{x + 4}$

Step 4

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

$x - 4$

Step 5

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

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Substitute $4$ for $x$ in $\frac{3}{x + 4}$ and simplify.

Substitute $4$ for $x$ to find the $y$ coordinate of the hole.

$\frac{3}{4 + 4}$

Add $4$ and $4$ .

$\frac{3}{8}$

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

$(4, \frac{3}{8})$

Step 6