Find the Holes in the Graph f(x)=(x^2+2x+1)/(x+1)

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

Step 1

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

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

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

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

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

Step 2

Cancel the common factor of ${(x + 1)}^{2}$ and $x + 1$ .

Factor $x + 1$ out of ${(x + 1)}^{2}$ .

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

Cancel the common factors.

Multiply by $1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $x + 1$ by $1$ .

$f (x) = x + 1$

Step 3

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

$x + 1$

Step 4

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $x + 1$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Substitute $- 1$ for $x$ in $x + 1$ and simplify.

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

$- 1 + 1$

Add $- 1$ and $1$ .

$0$

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

$(- 1, 0)$

Step 5