Find the Holes in the Graph (4x^2-4x-24)/(3x^2-10x+3)

$\frac{4 x^{2} - 4 x - 24}{3 x^{2} - 10 x + 3}$

Step 1

Factor $4 x^{2} - 4 x - 24$ .

Factor $4$ out of $4 x^{2} - 4 x - 24$ .

Factor $4$ out of $4 x^{2}$ .

$\frac{4 (x^{2}) - 4 x - 24}{3 x^{2} - 10 x + 3}$

Factor $4$ out of $- 4 x$ .

$\frac{4 (x^{2}) + 4 (- x) - 24}{3 x^{2} - 10 x + 3}$

Factor $4$ out of $- 24$ .

$\frac{4 (x^{2}) + 4 (- x) + 4 (- 6)}{3 x^{2} - 10 x + 3}$

Factor $4$ out of $4 (x^{2}) + 4 (- x)$ .

$\frac{4 (x^{2} - x) + 4 (- 6)}{3 x^{2} - 10 x + 3}$

Factor $4$ out of $4 (x^{2} - x) + 4 (- 6)$ .

$\frac{4 (x^{2} - x - 6)}{3 x^{2} - 10 x + 3}$

Factor.

Factor $x^{2} - x - 6$ 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

$\frac{4 ((x - 3) (x + 2))}{3 x^{2} - 10 x + 3}$

Remove unnecessary parentheses.

$\frac{4 (x - 3) (x + 2)}{3 x^{2} - 10 x + 3}$

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 $a \cdot c = 3 \cdot 3 = 9$ and whose sum is $b = - 10$ .

Factor $- 10$ out of $- 10 x$ .

$\frac{4 (x - 3) (x + 2)}{3 x^{2} - 10 (x) + 3}$

Rewrite $- 10$ as $- 1$ plus $- 9$

$\frac{4 (x - 3) (x + 2)}{3 x^{2} + (- 1 - 9) x + 3}$

Apply the distributive property.

$\frac{4 (x - 3) (x + 2)}{3 x^{2} - 1 x - 9 x + 3}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{4 (x - 3) (x + 2)}{(3 x^{2} - 1 x) - 9 x + 3}$

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

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

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

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

Step 3

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$\frac{4 (3 + 2)}{3 \cdot 3 - 1}$

Simplify.

Add $3$ and $2$ .

$\frac{4 \cdot 5}{3 \cdot 3 - 1}$

Simplify the denominator.

Multiply $3$ by $3$ .

$\frac{4 \cdot 5}{9 - 1}$

Subtract $1$ from $9$ .

$\frac{4 \cdot 5}{8}$

Multiply $4$ by $5$ .

$\frac{20}{8}$

Cancel the common factor of $20$ and $8$ .

Factor $4$ out of $20$ .

$\frac{4 (5)}{8}$

Cancel the common factors.

Factor $4$ out of $8$ .

$\frac{4 \cdot 5}{4 \cdot 2}$

Cancel the common factor.

$\frac{4 \cdot 5}{4 \cdot 2}$

Rewrite the expression.

$\frac{5}{2}$

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

$(3, \frac{5}{2})$

Step 6

Find the Holes in the Graph (4x^2-4x-24)/(3x^2-10x+3)

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