Find the Holes in the Graph f(x)=(2x^2-5x-12)/(3x^2-11x-4)

$f (x) = \frac{2 x^{2} - 5 x - 12}{3 x^{2} - 11 x - 4}$

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 $a \cdot c = 2 \cdot - 12 = - 24$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$f (x) = \frac{2 x^{2} - 5 x - 12}{3 x^{2} - 11 x - 4}$

Rewrite $- 5$ as $3$ plus $- 8$

$f (x) = \frac{2 x^{2} + (3 - 8) x - 12}{3 x^{2} - 11 x - 4}$

Apply the distributive property.

$f (x) = \frac{2 x^{2} + 3 x - 8 x - 12}{3 x^{2} - 11 x - 4}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(2 x^{2} + 3 x) - 8 x - 12}{3 x^{2} - 11 x - 4}$

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

$f (x) = \frac{x (2 x + 3) - 4 (2 x + 3)}{3 x^{2} - 11 x - 4}$

Factor the polynomial by factoring out the greatest common factor, $2 x + 3$ .

$f (x) = \frac{(2 x + 3) (x - 4)}{3 x^{2} - 11 x - 4}$

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 $a \cdot c = 3 \cdot - 4 = - 12$ and whose sum is $b = - 11$ .

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

$f (x) = \frac{(2 x + 3) (x - 4)}{3 x^{2} - 11 x - 4}$

Rewrite $- 11$ as $1$ plus $- 12$

$f (x) = \frac{(2 x + 3) (x - 4)}{3 x^{2} + (1 - 12) x - 4}$

Apply the distributive property.

$f (x) = \frac{(2 x + 3) (x - 4)}{3 x^{2} + 1 x - 12 x - 4}$

Multiply $x$ by $1$ .

$f (x) = \frac{(2 x + 3) (x - 4)}{3 x^{2} + x - 12 x - 4}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(2 x + 3) (x - 4)}{(3 x^{2} + x) - 12 x - 4}$

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

$f (x) = \frac{(2 x + 3) (x - 4)}{x (3 x + 1) - 4 (3 x + 1)}$

Factor the polynomial by factoring out the greatest common factor, $3 x + 1$ .

$f (x) = \frac{(2 x + 3) (x - 4)}{(3 x + 1) (x - 4)}$

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$f (x) = \frac{(2 x + 3) (x - 4)}{(3 x + 1) (x - 4)}$

Rewrite the expression.

$f (x) = \frac{2 x + 3}{3 x + 1}$

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

$x - 4$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{2 x + 3}{3 x + 1}$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Substitute $4$ for $x$ in $\frac{2 x + 3}{3 x + 1}$ and simplify.

Substitute $4$ for $x$ to find the $y$ coordinate of the hole.

$\frac{2 \cdot 4 + 3}{3 \cdot 4 + 1}$

Simplify.

Simplify the numerator.

Multiply $2$ by $4$ .

$\frac{8 + 3}{3 \cdot 4 + 1}$

Add $8$ and $3$ .

$\frac{11}{3 \cdot 4 + 1}$

Simplify the denominator.

Multiply $3$ by $4$ .

$\frac{11}{12 + 1}$

Add $12$ and $1$ .

$\frac{11}{13}$

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

$(4, \frac{11}{13})$

Find the Holes in the Graph f(x)=(2x^2-5x-12)/(3x^2-11x-4)

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