Find the Holes in the Graph f(x)=(x^2+8x-20)/(2x^2+20x)

$f (x) = \frac{x^{2} + 8 x - 20}{2 x^{2} + 20 x}$

Factor $x^{2} + 8 x - 20$ 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 $8$ .

$- 2, 10$

Write the factored form using these integers.

$f (x) = \frac{(x - 2) (x + 10)}{2 x^{2} + 20 x}$

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

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

$f (x) = \frac{(x - 2) (x + 10)}{2 x (x) + 20 x}$

Factor $2 x$ out of $20 x$ .

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

Factor $2 x$ out of $2 x (x) + 2 x (10)$ .

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

Cancel the common factor of $x + 10$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x + 10$

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

Set $x + 10$ equal to $0$ .

$x + 10 = 0$

Subtract $10$ from both sides of the equation.

$x = - 10$

Substitute $- 10$ for $x$ in $\frac{x - 2}{2 x}$ and simplify.

Substitute $- 10$ for $x$ to find the $y$ coordinate of the hole.

$\frac{- 10 - 2}{2 \cdot - 10}$

Simplify.

Cancel the common factor of $- 10 - 2$ and $2$ .

Factor $2$ out of $- 10$ .

$\frac{2 \cdot - 5 - 2}{2 \cdot - 10}$

Factor $2$ out of $- 2$ .

$\frac{2 \cdot - 5 + 2 \cdot - 1}{2 \cdot - 10}$

Factor $2$ out of $2 \cdot - 5 + 2 \cdot - 1$ .

$\frac{2 \cdot (- 5 - 1)}{2 \cdot - 10}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 10$ .

$\frac{2 \cdot (- 5 - 1)}{2 (- 10)}$

Cancel the common factor.

$\frac{2 \cdot (- 5 - 1)}{2 \cdot - 10}$

Rewrite the expression.

$\frac{- 5 - 1}{- 10}$

Subtract $1$ from $- 5$ .

$\frac{- 6}{- 10}$

Cancel the common factor of $- 6$ and $- 10$ .

Factor $- 2$ out of $- 6$ .

$\frac{- 2 (3)}{- 10}$

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{3}{5}$

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

$(- 10, \frac{3}{5})$

Find the Holes in the Graph f(x)=(x^2+8x-20)/(2x^2+20x)

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