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

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

Factor $2$ out of $2 x + 2$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 (x) + 2 (1)$ .

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

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

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

$\frac{2 (x + 1)}{x^{2} x + x^{2}}$

Multiply by $1$ .

$\frac{2 (x + 1)}{x^{2} x + x^{2} \cdot 1}$

Factor $x^{2}$ out of $x^{2} x + x^{2} \cdot 1$ .

$\frac{2 (x + 1)}{x^{2} (x + 1)}$

Cancel the common factor of $x + 1$ .

Cancel the common factor.

$\frac{2 (x + 1)}{x^{2} (x + 1)}$

Rewrite the expression.

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

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

$x + 1$

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

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Substitute $- 1$ for $x$ in $\frac{2}{x^{2}}$ and simplify.

Substitute $- 1$ for $x$ to find the $y$ coordinate of the hole.

$\frac{2}{{(- 1)}^{2}}$

Simplify.

Raise $- 1$ to the power of $2$ .

$\frac{2}{1}$

Divide $2$ by $1$ .

$2$

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

$(- 1, 2)$