Find the Holes in the Graph f(x)=(3x^2-7x-6)/(2x^2-5x-3)

$f (x) = \frac{3 x^{2} - 7 x - 6}{2 x^{2} - 5 x - 3}$

Step 1

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 $a \cdot c = 3 \cdot - 6 = - 18$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$f (x) = \frac{3 x^{2} - 7 x - 6}{2 x^{2} - 5 x - 3}$

Rewrite $- 7$ as $2$ plus $- 9$

$f (x) = \frac{3 x^{2} + (2 - 9) x - 6}{2 x^{2} - 5 x - 3}$

Apply the distributive property.

$f (x) = \frac{3 x^{2} + 2 x - 9 x - 6}{2 x^{2} - 5 x - 3}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(3 x^{2} + 2 x) - 9 x - 6}{2 x^{2} - 5 x - 3}$

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

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

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

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

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 $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = - 5$ .

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

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

Rewrite $- 5$ as $1$ plus $- 6$

$f (x) = \frac{(3 x + 2) (x - 3)}{2 x^{2} + (1 - 6) x - 3}$

Apply the distributive property.

$f (x) = \frac{(3 x + 2) (x - 3)}{2 x^{2} + 1 x - 6 x - 3}$

Multiply $x$ by $1$ .

$f (x) = \frac{(3 x + 2) (x - 3)}{2 x^{2} + x - 6 x - 3}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(3 x + 2) (x - 3)}{(2 x^{2} + x) - 6 x - 3}$

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

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

Step 3

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x - 3$

Step 5

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

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

Substitute $3$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Simplify the numerator.

Multiply $3$ by $3$ .

$\frac{9 + 2}{2 \cdot 3 + 1}$

Add $9$ and $2$ .

$\frac{11}{2 \cdot 3 + 1}$

Simplify the denominator.

Multiply $2$ by $3$ .

$\frac{11}{6 + 1}$

Add $6$ and $1$ .

$\frac{11}{7}$

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

$(3, \frac{11}{7})$

Step 6

Find the Holes in the Graph f(x)=(3x^2-7x-6)/(2x^2-5x-3)

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