Find the Holes in the Graph f(x)=(2x^3+8x^2-10x)/(2x^2+2x)

$f (x) = \frac{2 x^{3} + 8 x^{2} - 10 x}{2 x^{2} + 2 x}$

Factor $2 x^{3} + 8 x^{2} - 10 x$ .

Factor $2 x$ out of $2 x^{3} + 8 x^{2} - 10 x$ .

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

$f (x) = \frac{2 x (x^{2}) + 8 x^{2} - 10 x}{2 x^{2} + 2 x}$

Factor $2 x$ out of $8 x^{2}$ .

$f (x) = \frac{2 x (x^{2}) + 2 x (4 x) - 10 x}{2 x^{2} + 2 x}$

Factor $2 x$ out of $- 10 x$ .

$f (x) = \frac{2 x (x^{2}) + 2 x (4 x) + 2 x (- 5)}{2 x^{2} + 2 x}$

Factor $2 x$ out of $2 x (x^{2}) + 2 x (4 x)$ .

$f (x) = \frac{2 x (x^{2} + 4 x) + 2 x (- 5)}{2 x^{2} + 2 x}$

Factor $2 x$ out of $2 x (x^{2} + 4 x) + 2 x (- 5)$ .

$f (x) = \frac{2 x (x^{2} + 4 x - 5)}{2 x^{2} + 2 x}$

Factor.

Factor $x^{2} + 4 x - 5$ 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 $- 5$ and whose sum is $4$ .

$- 1, 5$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

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

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

Factor $2 x$ out of $2 x$ .

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

Factor $2 x$ out of $2 x (x) + 2 x (1)$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{(x - 1) (x + 5)}{x + 1}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{(x - 1) (x + 5)}{x + 1}$ and simplify.

Substitute $0$ for $x$ to find the $y$ coordinate of the hole.

$\frac{(0 - 1) (0 + 5)}{0 + 1}$

Simplify.

Cancel the common factor of $0 - 1$ and $0 + 1$ .

Rewrite $0$ as $- 1 (0)$ .

$\frac{(- 1 (0) - 1) (0 + 5)}{0 + 1}$

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

$\frac{(- 1 \cdot 0 - 1 \cdot 1) (0 + 5)}{0 + 1}$

Factor $- 1$ out of $- 1 \cdot 0 - 1 \cdot 1$ .

$\frac{- 1 \cdot (0 + 1) (0 + 5)}{0 + 1}$

Cancel the common factor.

$\frac{- 1 \cdot (0 + 1) (0 + 5)}{0 + 1}$

Divide $- 1 (0 + 5)$ by $1$ .

$- 1 (0 + 5)$

Rewrite $- 1 (0 + 5)$ as $- (0 + 5)$ .

$- (0 + 5)$

Add $0$ and $5$ .

$- 1 \cdot 5$

Multiply $- 1$ by $5$ .

$- 5$

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

$(0, - 5)$

Find the Holes in the Graph f(x)=(2x^3+8x^2-10x)/(2x^2+2x)

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