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

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

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of ${(x + 3)}^{2}$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{- 2 (x + 2)}{3 (x + 2)}$

Cancel the common factor of $x + 2$ .

Cancel the common factor.

$\frac{- 2 (x + 2)}{3 (x + 2)}$

Rewrite the expression.

$\frac{- 2}{3}$

Move the negative in front of the fraction.

$- \frac{2}{3}$

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

$x, x + 2, x + 3$

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

Set $x$ equal to $0$ .

$x = 0$

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Because the cancelled expression is constant with respect to $x$ , it is the $y$ coordinate for every hole.

$- \frac{2}{3}$

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

$(0, - \frac{2}{3}), (- 2, - \frac{2}{3}), (- 3, - \frac{2}{3})$