Find the Holes in the Graph f(x)=(2x^2+9x-18)/(x^2+7x+6)

$f (x) = \frac{2 x^{2} + 9 x - 18}{x^{2} + 7 x + 6}$

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 $a \cdot c = 2 \cdot - 18 = - 36$ and whose sum is $b = 9$ .

Factor $9$ out of $9 x$ .

$f (x) = \frac{2 x^{2} + 9 (x) - 18}{x^{2} + 7 x + 6}$

Rewrite $9$ as $- 3$ plus $12$

$f (x) = \frac{2 x^{2} + (- 3 + 12) x - 18}{x^{2} + 7 x + 6}$

Apply the distributive property.

$f (x) = \frac{2 x^{2} - 3 x + 12 x - 18}{x^{2} + 7 x + 6}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(2 x^{2} - 3 x) + 12 x - 18}{x^{2} + 7 x + 6}$

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

$f (x) = \frac{x (2 x - 3) + 6 (2 x - 3)}{x^{2} + 7 x + 6}$

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

$f (x) = \frac{(2 x - 3) (x + 6)}{x^{2} + 7 x + 6}$

Factor $x^{2} + 7 x + 6$ 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 $7$ .

$1, 6$

Write the factored form using these integers.

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

Cancel the common factor of $x + 6$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x + 6$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{2 x - 3}{x + 1}$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Substitute $- 6$ for $x$ in $\frac{2 x - 3}{x + 1}$ and simplify.

Substitute $- 6$ for $x$ to find the $y$ coordinate of the hole.

$\frac{2 \cdot - 6 - 3}{- 6 + 1}$

Simplify.

Simplify the numerator.

Multiply $2$ by $- 6$ .

$\frac{- 12 - 3}{- 6 + 1}$

Subtract $3$ from $- 12$ .

$\frac{- 15}{- 6 + 1}$

Add $- 6$ and $1$ .

$\frac{- 15}{- 5}$

Divide $- 15$ by $- 5$ .

$3$

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

$(- 6, 3)$

Find the Holes in the Graph f(x)=(2x^2+9x-18)/(x^2+7x+6)

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