Find the Holes in the Graph f(x)=(3x^2+26x+16)/(2x+16)

$f (x) = \frac{3 x^{2} + 26 x + 16}{2 x + 16}$

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 $b = 26$ .

Factor $26$ out of $26 x$ .

$f (x) = \frac{3 x^{2} + 26 (x) + 16}{2 x + 16}$

Rewrite $26$ as $2$ plus $24$

$f (x) = \frac{3 x^{2} + (2 + 24) x + 16}{2 x + 16}$

Apply the distributive property.

$f (x) = \frac{3 x^{2} + 2 x + 24 x + 16}{2 x + 16}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(3 x^{2} + 2 x) + 24 x + 16}{2 x + 16}$

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

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

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

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

Factor $2$ out of $2 x + 16$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $16$ .

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

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

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

Cancel the common factor of $x + 8$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x + 8$

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

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

Substitute $- 8$ for $x$ in $\frac{3 x + 2}{2}$ and simplify.

Substitute $- 8$ for $x$ to find the $y$ coordinate of the hole.

$\frac{3 \cdot - 8 + 2}{2}$

Simplify.

Cancel the common factor of $3 \cdot - 8 + 2$ and $2$ .

Factor $2$ out of $3 \cdot - 8$ .

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 (3 \cdot - 4) + 2 (1)$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $3 \cdot - 4 + 1$ by $1$ .

$3 \cdot - 4 + 1$

Multiply $3$ by $- 4$ .

$- 12 + 1$

Add $- 12$ and $1$ .

$- 11$

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

$(- 8, - 11)$

Find the Holes in the Graph f(x)=(3x^2+26x+16)/(2x+16)

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