Find the Holes in the Graph f(x)=(2x+14)/(3x^2+23x+14)

$f (x) = \frac{2 x + 14}{3 x^{2} + 23 x + 14}$

Step 1

Factor $2$ out of $2 x + 14$ .

Factor $2$ out of $2 x$ .

$f (x) = \frac{2 (x) + 14}{3 x^{2} + 23 x + 14}$

Factor $2$ out of $14$ .

$f (x) = \frac{2 x + 2 \cdot 7}{3 x^{2} + 23 x + 14}$

Factor $2$ out of $2 x + 2 \cdot 7$ .

$f (x) = \frac{2 (x + 7)}{3 x^{2} + 23 x + 14}$

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 $a \cdot c = 3 \cdot 14 = 42$ and whose sum is $b = 23$ .

Factor $23$ out of $23 x$ .

$f (x) = \frac{2 (x + 7)}{3 x^{2} + 23 (x) + 14}$

Rewrite $23$ as $2$ plus $21$

$f (x) = \frac{2 (x + 7)}{3 x^{2} + (2 + 21) x + 14}$

Apply the distributive property.

$f (x) = \frac{2 (x + 7)}{3 x^{2} + 2 x + 21 x + 14}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{2 (x + 7)}{(3 x^{2} + 2 x) + 21 x + 14}$

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

$f (x) = \frac{2 (x + 7)}{x (3 x + 2) + 7 (3 x + 2)}$

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

$f (x) = \frac{2 (x + 7)}{(3 x + 2) (x + 7)}$

Step 3

Cancel the common factor of $x + 7$ .

Cancel the common factor.

$f (x) = \frac{2 (x + 7)}{(3 x + 2) (x + 7)}$

Rewrite the expression.

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

Step 4

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

$x + 7$

Step 5

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

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Substitute $- 7$ for $x$ in $\frac{2}{3 x + 2}$ and simplify.

Substitute $- 7$ for $x$ to find the $y$ coordinate of the hole.

$\frac{2}{3 \cdot - 7 + 2}$

Simplify.

Simplify the denominator.

Multiply $3$ by $- 7$ .

$\frac{2}{- 21 + 2}$

Add $- 21$ and $2$ .

$\frac{2}{- 19}$

Move the negative in front of the fraction.

$- \frac{2}{19}$

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

$(- 7, - \frac{2}{19})$

Step 6

Find the Holes in the Graph f(x)=(2x+14)/(3x^2+23x+14)

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