Find the Holes in the Graph f(x)=(2x^2-2x-40)/(2x^2-14x+20)

$f (x) = \frac{2 x^{2} - 2 x - 40}{2 x^{2} - 14 x + 20}$

Factor $2 x^{2} - 2 x - 40$ .

Factor $2$ out of $2 x^{2} - 2 x - 40$ .

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

$f (x) = \frac{2 (x^{2}) - 2 x - 40}{2 x^{2} - 14 x + 20}$

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

$f (x) = \frac{2 (x^{2}) + 2 (- x) - 40}{2 x^{2} - 14 x + 20}$

Factor $2$ out of $- 40$ .

$f (x) = \frac{2 (x^{2}) + 2 (- x) + 2 (- 20)}{2 x^{2} - 14 x + 20}$

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

$f (x) = \frac{2 (x^{2} - x) + 2 (- 20)}{2 x^{2} - 14 x + 20}$

Factor $2$ out of $2 (x^{2} - x) + 2 (- 20)$ .

$f (x) = \frac{2 (x^{2} - x - 20)}{2 x^{2} - 14 x + 20}$

Factor.

Factor $x^{2} - x - 20$ 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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Write the factored form using these integers.

$f (x) = \frac{2 ((x - 5) (x + 4))}{2 x^{2} - 14 x + 20}$

Remove unnecessary parentheses.

$f (x) = \frac{2 (x - 5) (x + 4)}{2 x^{2} - 14 x + 20}$

Factor $2 x^{2} - 14 x + 20$ .

Factor $2$ out of $2 x^{2} - 14 x + 20$ .

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

$f (x) = \frac{2 (x - 5) (x + 4)}{2 (x^{2}) - 14 x + 20}$

Factor $2$ out of $- 14 x$ .

$f (x) = \frac{2 (x - 5) (x + 4)}{2 (x^{2}) + 2 (- 7 x) + 20}$

Factor $2$ out of $20$ .

$f (x) = \frac{2 (x - 5) (x + 4)}{2 x^{2} + 2 (- 7 x) + 2 \cdot 10}$

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

$f (x) = \frac{2 (x - 5) (x + 4)}{2 (x^{2} - 7 x) + 2 \cdot 10}$

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

$f (x) = \frac{2 (x - 5) (x + 4)}{2 (x^{2} - 7 x + 10)}$

Factor.

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 $- 7$ .

$- 5, - 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 5$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{x + 4}{x - 2}$ .

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Substitute $5$ for $x$ in $\frac{x + 4}{x - 2}$ and simplify.

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

$\frac{5 + 4}{5 - 2}$

Simplify.

Add $5$ and $4$ .

$\frac{9}{5 - 2}$

Subtract $2$ from $5$ .

$\frac{9}{3}$

Divide $9$ by $3$ .

$3$

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

$(5, 3)$

Find the Holes in the Graph f(x)=(2x^2-2x-40)/(2x^2-14x+20)

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