Find the Holes in the Graph f(x)=(3x-15)/(3x^2-11x-20)

$f (x) = \frac{3 x - 15}{3 x^{2} - 11 x - 20}$

Factor $3$ out of $3 x - 15$ .

Factor $3$ out of $3 x$ .

$f (x) = \frac{3 (x) - 15}{3 x^{2} - 11 x - 20}$

Factor $3$ out of $- 15$ .

$f (x) = \frac{3 x + 3 \cdot - 5}{3 x^{2} - 11 x - 20}$

Factor $3$ out of $3 x + 3 \cdot - 5$ .

$f (x) = \frac{3 (x - 5)}{3 x^{2} - 11 x - 20}$

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 $a \cdot c = 3 \cdot - 20 = - 60$ and whose sum is $b = - 11$ .

Factor $- 11$ out of $- 11 x$ .

$f (x) = \frac{3 (x - 5)}{3 x^{2} - 11 x - 20}$

Rewrite $- 11$ as $4$ plus $- 15$

$f (x) = \frac{3 (x - 5)}{3 x^{2} + (4 - 15) x - 20}$

Apply the distributive property.

$f (x) = \frac{3 (x - 5)}{3 x^{2} + 4 x - 15 x - 20}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{3 (x - 5)}{(3 x^{2} + 4 x) - 15 x - 20}$

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x + 4$ .

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

Cancel the common factor of $x - 5$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{3}{3 x + 4}$ .

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

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

$\frac{3}{3 \cdot 5 + 4}$

Simplify the denominator.

Multiply $3$ by $5$ .

$\frac{3}{15 + 4}$

Add $15$ and $4$ .

$\frac{3}{19}$

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

$(5, \frac{3}{19})$

Find the Holes in the Graph f(x)=(3x-15)/(3x^2-11x-20)

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