Find the Holes in the Graph y=(x^2+5x)/(x^2+7x+10)

$y = \frac{x^{2} + 5 x}{x^{2} + 7 x + 10}$

Factor $x$ out of $x^{2} + 5 x$ .

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

$y = \frac{x \cdot x + 5 x}{x^{2} + 7 x + 10}$

Factor $x$ out of $5 x$ .

$y = \frac{x \cdot x + x \cdot 5}{x^{2} + 7 x + 10}$

Factor $x$ out of $x \cdot x + x \cdot 5$ .

$y = \frac{x (x + 5)}{x^{2} + 7 x + 10}$

Factor $x^{2} + 7 x + 10$ 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 $10$ and whose sum is $7$ .

$2, 5$

Write the factored form using these integers.

$y = \frac{x (x + 5)}{(x + 2) (x + 5)}$

Cancel the common factor of $x + 5$ .

Cancel the common factor.

$y = \frac{x (x + 5)}{(x + 2) (x + 5)}$

Rewrite the expression.

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

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

$x + 5$

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 $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Substitute $- 5$ for $x$ in $\frac{x}{x + 2}$ and simplify.

Substitute $- 5$ for $x$ to find the $y$ coordinate of the hole.

$\frac{- 5}{- 5 + 2}$

Simplify.

Add $- 5$ and $2$ .

$\frac{- 5}{- 3}$

Dividing two negative values results in a positive value.

$\frac{5}{3}$

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

$(- 5, \frac{5}{3})$

Find the Holes in the Graph y=(x^2+5x)/(x^2+7x+10)

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