Find the Holes in the Graph (5x^2+20x)/(3x+12)

$\frac{5 x^{2} + 20 x}{3 x + 12}$

Factor $5 x$ out of $5 x^{2} + 20 x$ .

Factor $5 x$ out of $5 x^{2}$ .

$\frac{5 x (x) + 20 x}{3 x + 12}$

Factor $5 x$ out of $20 x$ .

$\frac{5 x (x) + 5 x (4)}{3 x + 12}$

Factor $5 x$ out of $5 x (x) + 5 x (4)$ .

$\frac{5 x (x + 4)}{3 x + 12}$

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

Factor $3$ out of $3 x$ .

$\frac{5 x (x + 4)}{3 (x) + 12}$

Factor $3$ out of $12$ .

$\frac{5 x (x + 4)}{3 x + 3 \cdot 4}$

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

$\frac{5 x (x + 4)}{3 (x + 4)}$

Cancel the common factor of $x + 4$ .

Cancel the common factor.

$\frac{5 x (x + 4)}{3 (x + 4)}$

Rewrite the expression.

$\frac{5 x}{3}$

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

$x + 4$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{5 x}{3}$ .

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Substitute $- 4$ for $x$ in $\frac{5 x}{3}$ and simplify.

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

$\frac{5 \cdot - 4}{3}$

Simplify.

Multiply $5$ by $- 4$ .

$\frac{- 20}{3}$

Move the negative in front of the fraction.

$- \frac{20}{3}$

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

$(- 4, - \frac{20}{3})$