Find the Holes in the Graph f(x)=(2x^2+18x+40)/(3x^2+10x-25)

$f (x) = \frac{2 x^{2} + 18 x + 40}{3 x^{2} + 10 x - 25}$

Step 1

Factor $2 x^{2} + 18 x + 40$ .

Factor $2$ out of $2 x^{2} + 18 x + 40$ .

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

$f (x) = \frac{2 (x^{2}) + 18 x + 40}{3 x^{2} + 10 x - 25}$

Factor $2$ out of $18 x$ .

$f (x) = \frac{2 (x^{2}) + 2 (9 x) + 40}{3 x^{2} + 10 x - 25}$

Factor $2$ out of $40$ .

$f (x) = \frac{2 x^{2} + 2 (9 x) + 2 \cdot 20}{3 x^{2} + 10 x - 25}$

Factor $2$ out of $2 x^{2} + 2 (9 x)$ .

$f (x) = \frac{2 (x^{2} + 9 x) + 2 \cdot 20}{3 x^{2} + 10 x - 25}$

Factor $2$ out of $2 (x^{2} + 9 x) + 2 \cdot 20$ .

$f (x) = \frac{2 (x^{2} + 9 x + 20)}{3 x^{2} + 10 x - 25}$

Factor.

Factor $x^{2} + 9 x + 20$ 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 $9$ .

$4, 5$

Write the factored form using these integers.

$f (x) = \frac{2 ((x + 4) (x + 5))}{3 x^{2} + 10 x - 25}$

Remove unnecessary parentheses.

$f (x) = \frac{2 (x + 4) (x + 5)}{3 x^{2} + 10 x - 25}$

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 $a \cdot c = 3 \cdot - 25 = - 75$ and whose sum is $b = 10$ .

Factor $10$ out of $10 x$ .

$f (x) = \frac{2 (x + 4) (x + 5)}{3 x^{2} + 10 (x) - 25}$

Rewrite $10$ as $- 5$ plus $15$

$f (x) = \frac{2 (x + 4) (x + 5)}{3 x^{2} + (- 5 + 15) x - 25}$

Apply the distributive property.

$f (x) = \frac{2 (x + 4) (x + 5)}{3 x^{2} - 5 x + 15 x - 25}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{2 (x + 4) (x + 5)}{(3 x^{2} - 5 x) + 15 x - 25}$

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

$f (x) = \frac{2 (x + 4) (x + 5)}{x (3 x - 5) + 5 (3 x - 5)}$

Factor the polynomial by factoring out the greatest common factor, $3 x - 5$ .

$f (x) = \frac{2 (x + 4) (x + 5)}{(3 x - 5) (x + 5)}$

Step 3

Cancel the common factor of $x + 5$ .

Cancel the common factor.

$f (x) = \frac{2 (x + 4) (x + 5)}{(3 x - 5) (x + 5)}$

Rewrite the expression.

$f (x) = \frac{2 (x + 4)}{3 x - 5}$

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 $\frac{2 (x + 4)}{3 x - 5}$ .

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Substitute $- 5$ for $x$ in $\frac{2 (x + 4)}{3 x - 5}$ and simplify.

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

$\frac{2 (- 5 + 4)}{3 \cdot - 5 - 5}$

Simplify.

Add $- 5$ and $4$ .

$\frac{2 \cdot - 1}{3 \cdot - 5 - 5}$

Simplify the denominator.

Multiply $3$ by $- 5$ .

$\frac{2 \cdot - 1}{- 15 - 5}$

Subtract $5$ from $- 15$ .

$\frac{2 \cdot - 1}{- 20}$

Multiply $2$ by $- 1$ .

$\frac{- 2}{- 20}$

Cancel the common factor of $- 2$ and $- 20$ .

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

$\frac{- 2 (1)}{- 20}$

Cancel the common factors.

Factor $- 2$ out of $- 20$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{10}$

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

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

Step 6

Find the Holes in the Graph f(x)=(2x^2+18x+40)/(3x^2+10x-25)

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