Find the Holes in the Graph (x^2+3x)/(x^2+5x+6)

$\frac{x^{2} + 3 x}{x^{2} + 5 x + 6}$

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

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

$\frac{x \cdot x + 3 x}{x^{2} + 5 x + 6}$

Factor $x$ out of $3 x$ .

$\frac{x \cdot x + x \cdot 3}{x^{2} + 5 x + 6}$

Factor $x$ out of $x \cdot x + x \cdot 3$ .

$\frac{x (x + 3)}{x^{2} + 5 x + 6}$

Factor $x^{2} + 5 x + 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $5$ .

$2, 3$

Write the factored form using these integers.

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

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x + 3$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

Substitute $- 3$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Add $- 3$ and $2$ .

$\frac{- 3}{- 1}$

Divide $- 3$ by $- 1$ .

$3$

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

$(- 3, 3)$

Find the Holes in the Graph (x^2+3x)/(x^2+5x+6)

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