Find the Holes in the Graph (x^2+4x+4)/(x+2)

$\frac{x^{2} + 4 x + 4}{x + 2}$

Step 1

Factor using the perfect square rule.

Rewrite $4$ as $2^{2}$ .

$\frac{x^{2} + 4 x + 2^{2}}{x + 2}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 2$ .

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

Step 2

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

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

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

Cancel the common factors.

Multiply by $1$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{x + 2}{1}$

Divide $x + 2$ by $1$ .

$x + 2$

Step 3

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

$x + 2$

Step 4

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Substitute $- 2$ for $x$ in $x + 2$ and simplify.

Substitute $- 2$ for $x$ to find the $y$ coordinate of the hole.

$- 2 + 2$

Add $- 2$ and $2$ .

$0$

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

$(- 2, 0)$

Step 5