Find the Holes in the Graph f(x)=(-3x^2-15x)/(x^2+7x+10)

$f (x) = \frac{- 3 x^{2} - 15 x}{x^{2} + 7 x + 10}$

Factor $3 x$ out of $- 3 x^{2} - 15 x$ .

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

$f (x) = \frac{3 x (- x) - 15 x}{x^{2} + 7 x + 10}$

Factor $3 x$ out of $- 15 x$ .

$f (x) = \frac{3 x (- x) + 3 x (- 5)}{x^{2} + 7 x + 10}$

Factor $3 x$ out of $3 x (- x) + 3 x (- 5)$ .

$f (x) = \frac{3 x (- x - 5)}{x^{2} + 7 x + 10}$

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.

$f (x) = \frac{3 x (- x - 5)}{(x + 2) (x + 5)}$

Cancel the common factor of $- x - 5$ and $x + 5$ .

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

$f (x) = \frac{3 x (- (x) - 5)}{(x + 2) (x + 5)}$

Rewrite $- 5$ as $- 1 (5)$ .

$f (x) = \frac{3 x (- (x) - 1 \cdot 5)}{(x + 2) (x + 5)}$

Factor $- 1$ out of $- (x) - 1 (5)$ .

$f (x) = \frac{3 x (- (x + 5))}{(x + 2) (x + 5)}$

Rewrite $- (x + 5)$ as $- 1 (x + 5)$ .

$f (x) = \frac{3 x (- 1 (x + 5))}{(x + 2) (x + 5)}$

Cancel the common factor.

$f (x) = \frac{3 x (- 1 (x + 5))}{(x + 2) (x + 5)}$

Rewrite the expression.

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

Multiply $- 1$ by $3$ .

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

Move the negative in front of the fraction.

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

Multiply $3$ by $x$ .

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

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

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Substitute $- 5$ for $x$ in $- \frac{3 x}{x + 2}$ and simplify.

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

$- \frac{3 \cdot - 5}{- 5 + 2}$

Simplify.

Multiply $3$ by $- 5$ .

$- \frac{- 15}{- 5 + 2}$

Add $- 5$ and $2$ .

$- \frac{- 15}{- 3}$

Divide $- 15$ by $- 3$ .

$- 1 \cdot 5$

Multiply $- 1$ by $5$ .

$- 5$

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

$(- 5, - 5)$

Find the Holes in the Graph f(x)=(-3x^2-15x)/(x^2+7x+10)

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