Find the Holes in the Graph f(x)=(x^3-x^2-6x)/(-3x^2-3x+18)

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

Step 1

Factor $x^{3} - x^{2} - 6 x$ .

Factor $x$ out of $x^{3} - x^{2} - 6 x$ .

Factor $x$ out of $x^{3}$ .

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

Factor $x$ out of $- x^{2}$ .

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

Factor $x$ out of $- 6 x$ .

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

Factor $x$ out of $x \cdot x^{2} + x (- x)$ .

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

Factor $x$ out of $x (x^{2} - x) + x \cdot - 6$ .

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

Factor.

Factor $x^{2} - x - 6$ 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 $- 1$ .

$- 3, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 2

Factor $- 3 x^{2} - 3 x + 18$ .

Factor $3$ out of $- 3 x^{2} - 3 x + 18$ .

Factor $3$ out of $- 3 x^{2}$ .

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

Factor $3$ out of $- 3 x$ .

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

Factor $3$ out of $18$ .

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

Factor $3$ out of $3 (- x^{2}) + 3 (- x)$ .

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

Factor $3$ out of $3 (- x^{2} - x) + 3 (6)$ .

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

Factor.

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 $a \cdot c = - 1 \cdot 6 = - 6$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

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

Rewrite $- 1$ as $2$ plus $- 3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Remove unnecessary parentheses.

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

Step 3

Since no factors can be removed from the denominator, there are no holes in the graph.

No holes

Step 4

Find the Holes in the Graph f(x)=(x^3-x^2-6x)/(-3x^2-3x+18)

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