Find the Holes in the Graph f(x)=(x^2+2x+1)/(2x^2-x-3)

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

Factor using the perfect square rule.

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

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

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.

$f (x) = \frac{x^{2} + 2 \cdot x \cdot 1 + 1^{2}}{2 x^{2} - x - 3}$

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

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

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

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

Rewrite $- 1$ as $2$ plus $- 3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

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

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

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

$f (x) = \frac{(x + 1) (x + 1)}{(x + 1) (2 x - 3)}$

Cancel the common factors.

Cancel the common factor.

$f (x) = \frac{(x + 1) (x + 1)}{(x + 1) (2 x - 3)}$

Rewrite the expression.

$f (x) = \frac{x + 1}{2 x - 3}$

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 $\frac{x + 1}{2 x - 3}$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Substitute $- 1$ for $x$ in $\frac{x + 1}{2 x - 3}$ and simplify.

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

$\frac{- 1 + 1}{2 \cdot - 1 - 3}$

Simplify.

Add $- 1$ and $1$ .

$\frac{0}{2 \cdot - 1 - 3}$

Simplify the denominator.

Multiply $2$ by $- 1$ .

$\frac{0}{- 2 - 3}$

Subtract $3$ from $- 2$ .

$\frac{0}{- 5}$

Divide $0$ by $- 5$ .

$0$

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

$(- 1, 0)$

Find the Holes in the Graph f(x)=(x^2+2x+1)/(2x^2-x-3)

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