Find the Holes in the Graph y=(x^2+3x)/(x+3)

$y = \frac{x^{2} + 3 x}{x + 3}$

Step 1

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

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

$y = \frac{x \cdot x + 3 x}{x + 3}$

Factor $x$ out of $3 x$ .

$y = \frac{x \cdot x + x \cdot 3}{x + 3}$

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

$y = \frac{x (x + 3)}{x + 3}$

Step 2

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$y = \frac{x (x + 3)}{x + 3}$

Divide $x$ by $1$ .

$y = x$

Step 3

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

$x + 3$

Step 4

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

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Substitute $- 3$ for $x$ to find the $y$ coordinate of the hole.

$- 3$

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

$(- 3, - 3)$

Step 5