Find the Holes in the Graph (5x^2+25x)/(-3x^2-21x-30)

$\frac{5 x^{2} + 25 x}{- 3 x^{2} - 21 x - 30}$

Step 1

Factor $5 x$ out of $5 x^{2} + 25 x$ .

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

$\frac{5 x (x) + 25 x}{- 3 x^{2} - 21 x - 30}$

Factor $5 x$ out of $25 x$ .

$\frac{5 x (x) + 5 x (5)}{- 3 x^{2} - 21 x - 30}$

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

$\frac{5 x (x + 5)}{- 3 x^{2} - 21 x - 30}$

Step 2

Factor $- 3 x^{2} - 21 x - 30$ .

Factor $3$ out of $- 3 x^{2} - 21 x - 30$ .

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

$\frac{5 x (x + 5)}{3 (- x^{2}) - 21 x - 30}$

Factor $3$ out of $- 21 x$ .

$\frac{5 x (x + 5)}{3 (- x^{2}) + 3 (- 7 x) - 30}$

Factor $3$ out of $- 30$ .

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

Factor $3$ out of $3 (- x^{2}) + 3 (- 7 x)$ .

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

Factor $3$ out of $3 (- x^{2} - 7 x) + 3 (- 10)$ .

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

Factor.

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 $a \cdot c = - 1 \cdot - 10 = 10$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

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

Rewrite $- 7$ as $- 2$ plus $- 5$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $- x - 2$ .

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

Remove unnecessary parentheses.

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

Step 3

Cancel the common factor of $x + 5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x + 5$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{5 x}{3 (- x - 2)}$ .

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

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

$\frac{5 \cdot - 5}{3 (- - 5 - 2)}$

Simplify.

Multiply $5$ by $- 5$ .

$\frac{- 25}{3 (- - 5 - 2)}$

Simplify the denominator.

Multiply $- 1$ by $- 5$ .

$\frac{- 25}{3 (5 - 2)}$

Subtract $2$ from $5$ .

$\frac{- 25}{3 \cdot 3}$

Multiply $3$ by $3$ .

$\frac{- 25}{9}$

Move the negative in front of the fraction.

$- \frac{25}{9}$

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

$(- 5, - \frac{25}{9})$

Step 6

Find the Holes in the Graph (5x^2+25x)/(-3x^2-21x-30)

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