Find the Holes in the Graph f(x)=(3x^2+6x)/(x^2+x)

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

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

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

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

Factor $3 x$ out of $6 x$ .

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

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

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

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

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

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

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

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

Factor $x$ out of $x^{1}$ .

$f (x) = \frac{3 x (x + 2)}{x \cdot x + x \cdot 1}$

Factor $x$ out of $x \cdot x + x \cdot 1$ .

$f (x) = \frac{3 x (x + 2)}{x (x + 1)}$

Cancel the common factor of $x$ .

Cancel the common factor.

$f (x) = \frac{3 x (x + 2)}{x (x + 1)}$

Rewrite the expression.

$f (x) = \frac{3 (x + 2)}{x + 1}$

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

$x$

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

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{3 (x + 2)}{x + 1}$ and simplify.

Substitute $0$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Add $0$ and $2$ .

$\frac{3 \cdot 2}{0 + 1}$

Add $0$ and $1$ .

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

Multiply $3$ by $2$ .

$\frac{6}{1}$

Divide $6$ by $1$ .

$6$

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

$(0, 6)$