Find the Holes in the Graph f(x)=(x^2-100)/(2x^2+10x-100)

$f (x) = \frac{x^{2} - 100}{2 x^{2} + 10 x - 100}$

Factor $x^{2} - 100$ .

Rewrite $100$ as $10^{2}$ .

$f (x) = \frac{x^{2} - 10^{2}}{2 x^{2} + 10 x - 100}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 10$ .

$f (x) = \frac{(x + 10) (x - 10)}{2 x^{2} + 10 x - 100}$

Factor $2 x^{2} + 10 x - 100$ .

Factor $2$ out of $2 x^{2} + 10 x - 100$ .

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

$f (x) = \frac{(x + 10) (x - 10)}{2 (x^{2}) + 10 x - 100}$

Factor $2$ out of $10 x$ .

$f (x) = \frac{(x + 10) (x - 10)}{2 (x^{2}) + 2 (5 x) - 100}$

Factor $2$ out of $- 100$ .

$f (x) = \frac{(x + 10) (x - 10)}{2 x^{2} + 2 (5 x) + 2 \cdot - 50}$

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

$f (x) = \frac{(x + 10) (x - 10)}{2 (x^{2} + 5 x) + 2 \cdot - 50}$

Factor $2$ out of $2 (x^{2} + 5 x) + 2 \cdot - 50$ .

$f (x) = \frac{(x + 10) (x - 10)}{2 (x^{2} + 5 x - 50)}$

Factor.

Factor $x^{2} + 5 x - 50$ 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 $- 50$ and whose sum is $5$ .

$- 5, 10$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Cancel the common factor of $x + 10$ .

Cancel the common factor.

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

Rewrite the expression.

$f (x) = \frac{x - 10}{2 (x - 5)}$

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 $\frac{x - 10}{2 (x - 5)}$ .

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

$x + 10 = 0$

Subtract $10$ from both sides of the equation.

$x = - 10$

Substitute $- 10$ for $x$ in $\frac{x - 10}{2 (x - 5)}$ and simplify.

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

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

Simplify.

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

Factor $2$ out of $- 10$ .

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

Factor $2$ out of $- 10$ .

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

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

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $- 5 - 5$ and $- 10 - 5$ .

Factor $5$ out of $- 5$ .

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

Factor $5$ out of $- 5$ .

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

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

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

Cancel the common factors.

Factor $5$ out of $- 10$ .

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

Factor $5$ out of $- 5$ .

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

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

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{- 1 - 1}{- 2 - 1}$

Subtract $1$ from $- 1$ .

$\frac{- 2}{- 2 - 1}$

Subtract $1$ from $- 2$ .

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

Dividing two negative values results in a positive value.

$\frac{2}{3}$

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

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

Find the Holes in the Graph f(x)=(x^2-100)/(2x^2+10x-100)

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