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

$\frac{x^{2} - 16}{x - 4}$

Step 1

Factor $x^{2} - 16$ .

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

$\frac{x^{2} - 4^{2}}{x - 4}$

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 $b = 4$ .

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

Step 2

Cancel the common factor of $x - 4$ .

Cancel the common factor.

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

Divide $x + 4$ by $1$ .

$x + 4$

Step 3

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

$x - 4$

Step 4

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Substitute $4$ for $x$ in $x + 4$ and simplify.

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

$4 + 4$

Add $4$ and $4$ .

$8$

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

$(4, 8)$

Step 5