Find the Holes in the Graph f(x)=(3-14x-5x^2)/(3+7x+2x^2)

$f (x) = \frac{3 - 14 x - 5 x^{2}}{3 + 7 x + 2 x^{2}}$

Factor by grouping.

Reorder terms.

$f (x) = \frac{- 5 x^{2} - 14 x + 3}{3 + 7 x + 2 x^{2}}$

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 $a \cdot c = - 5 \cdot 3 = - 15$ and whose sum is $b = - 14$ .

Factor $- 14$ out of $- 14 x$ .

$f (x) = \frac{- 5 x^{2} - 14 x + 3}{3 + 7 x + 2 x^{2}}$

Rewrite $- 14$ as $1$ plus $- 15$

$f (x) = \frac{- 5 x^{2} + (1 - 15) x + 3}{3 + 7 x + 2 x^{2}}$

Apply the distributive property.

$f (x) = \frac{- 5 x^{2} + 1 x - 15 x + 3}{3 + 7 x + 2 x^{2}}$

Multiply $x$ by $1$ .

$f (x) = \frac{- 5 x^{2} + x - 15 x + 3}{3 + 7 x + 2 x^{2}}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(- 5 x^{2} + x) - 15 x + 3}{3 + 7 x + 2 x^{2}}$

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

$f (x) = \frac{x (- 5 x + 1) + 3 (- 5 x + 1)}{3 + 7 x + 2 x^{2}}$

Factor the polynomial by factoring out the greatest common factor, $- 5 x + 1$ .

$f (x) = \frac{(- 5 x + 1) (x + 3)}{3 + 7 x + 2 x^{2}}$

Factor by grouping.

Reorder terms.

$f (x) = \frac{(- 5 x + 1) (x + 3)}{2 x^{2} + 7 x + 3}$

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 $b = 7$ .

Factor $7$ out of $7 x$ .

$f (x) = \frac{(- 5 x + 1) (x + 3)}{2 x^{2} + 7 (x) + 3}$

Rewrite $7$ as $1$ plus $6$

$f (x) = \frac{(- 5 x + 1) (x + 3)}{2 x^{2} + (1 + 6) x + 3}$

Apply the distributive property.

$f (x) = \frac{(- 5 x + 1) (x + 3)}{2 x^{2} + 1 x + 6 x + 3}$

Multiply $x$ by $1$ .

$f (x) = \frac{(- 5 x + 1) (x + 3)}{2 x^{2} + x + 6 x + 3}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(- 5 x + 1) (x + 3)}{(2 x^{2} + x) + 6 x + 3}$

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

$f (x) = \frac{(- 5 x + 1) (x + 3)}{x (2 x + 1) + 3 (2 x + 1)}$

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

$f (x) = \frac{(- 5 x + 1) (x + 3)}{(2 x + 1) (x + 3)}$

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$f (x) = \frac{(- 5 x + 1) (x + 3)}{(2 x + 1) (x + 3)}$

Rewrite the expression.

$f (x) = \frac{- 5 x + 1}{2 x + 1}$

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

$x + 3$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{- 5 x + 1}{2 x + 1}$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Substitute $- 3$ for $x$ in $\frac{- 5 x + 1}{2 x + 1}$ and simplify.

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

$\frac{- 5 \cdot - 3 + 1}{2 \cdot - 3 + 1}$

Simplify.

Simplify the numerator.

Multiply $- 5$ by $- 3$ .

$\frac{15 + 1}{2 \cdot - 3 + 1}$

Add $15$ and $1$ .

$\frac{16}{2 \cdot - 3 + 1}$

Simplify the denominator.

Multiply $2$ by $- 3$ .

$\frac{16}{- 6 + 1}$

Add $- 6$ and $1$ .

$\frac{16}{- 5}$

Move the negative in front of the fraction.

$- \frac{16}{5}$

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

$(- 3, - \frac{16}{5})$

Find the Holes in the Graph f(x)=(3-14x-5x^2)/(3+7x+2x^2)

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