Find the Holes in the Graph f(x)=(2x+8)/(3x^2+16x+16)

$f (x) = \frac{2 x + 8}{3 x^{2} + 16 x + 16}$

Step 1

Factor $2$ out of $2 x + 8$ .

Factor $2$ out of $2 x$ .

$f (x) = \frac{2 (x) + 8}{3 x^{2} + 16 x + 16}$

Factor $2$ out of $8$ .

$f (x) = \frac{2 x + 2 \cdot 4}{3 x^{2} + 16 x + 16}$

Factor $2$ out of $2 x + 2 \cdot 4$ .

$f (x) = \frac{2 (x + 4)}{3 x^{2} + 16 x + 16}$

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 $a \cdot c = 3 \cdot 16 = 48$ and whose sum is $b = 16$ .

Factor $16$ out of $16 x$ .

$f (x) = \frac{2 (x + 4)}{3 x^{2} + 16 (x) + 16}$

Rewrite $16$ as $4$ plus $12$

$f (x) = \frac{2 (x + 4)}{3 x^{2} + (4 + 12) x + 16}$

Apply the distributive property.

$f (x) = \frac{2 (x + 4)}{3 x^{2} + 4 x + 12 x + 16}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{2 (x + 4)}{(3 x^{2} + 4 x) + 12 x + 16}$

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

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

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

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

Step 3

Cancel the common factor of $x + 4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x + 4$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{2}{3 x + 4}$ .

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

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

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

Simplify.

Cancel the common factor of $2$ and $3 \cdot - 4 + 4$ .

Factor $2$ out of $2$ .

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

Cancel the common factors.

Factor $2$ out of $3 \cdot - 4$ .

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

Factor $2$ out of $4$ .

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

Factor $2$ out of $2 (3 \cdot - 2) + 2 (2)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the denominator.

Multiply $3$ by $- 2$ .

$\frac{1}{- 6 + 2}$

Add $- 6$ and $2$ .

$\frac{1}{- 4}$

Move the negative in front of the fraction.

$- \frac{1}{4}$

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

$(- 4, - \frac{1}{4})$

Step 6

Find the Holes in the Graph f(x)=(2x+8)/(3x^2+16x+16)

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