Find the Holes in the Graph f(x)=(x^2-25)/(2x^2-17x+35)

$f (x) = \frac{x^{2} - 25}{2 x^{2} - 17 x + 35}$

Step 1

Factor $x^{2} - 25$ .

Rewrite $25$ as $5^{2}$ .

$f (x) = \frac{x^{2} - 5^{2}}{2 x^{2} - 17 x + 35}$

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 $b = 5$ .

$f (x) = \frac{(x + 5) (x - 5)}{2 x^{2} - 17 x + 35}$

Step 2

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 $a \cdot c = 2 \cdot 35 = 70$ and whose sum is $b = - 17$ .

Factor $- 17$ out of $- 17 x$ .

$f (x) = \frac{(x + 5) (x - 5)}{2 x^{2} - 17 x + 35}$

Rewrite $- 17$ as $- 7$ plus $- 10$

$f (x) = \frac{(x + 5) (x - 5)}{2 x^{2} + (- 7 - 10) x + 35}$

Apply the distributive property.

$f (x) = \frac{(x + 5) (x - 5)}{2 x^{2} - 7 x - 10 x + 35}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(x + 5) (x - 5)}{(2 x^{2} - 7 x) - 10 x + 35}$

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x - 7$ .

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

Step 3

Cancel the common factor of $x - 5$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{x + 5}{2 x - 7}$ .

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Substitute $5$ for $x$ in $\frac{x + 5}{2 x - 7}$ and simplify.

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

$\frac{5 + 5}{2 \cdot 5 - 7}$

Simplify.

Add $5$ and $5$ .

$\frac{10}{2 \cdot 5 - 7}$

Simplify the denominator.

Multiply $2$ by $5$ .

$\frac{10}{10 - 7}$

Subtract $7$ from $10$ .

$\frac{10}{3}$

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

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

Step 6

Find the Holes in the Graph f(x)=(x^2-25)/(2x^2-17x+35)

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