Find the Holes in the Graph f(x)=(x^3+5x^2+6x)/(3x^2+9x)

$f (x) = \frac{x^{3} + 5 x^{2} + 6 x}{3 x^{2} + 9 x}$

Step 1

Factor $x^{3} + 5 x^{2} + 6 x$ .

Factor $x$ out of $x^{3} + 5 x^{2} + 6 x$ .

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

$f (x) = \frac{x \cdot x^{2} + 5 x^{2} + 6 x}{3 x^{2} + 9 x}$

Factor $x$ out of $5 x^{2}$ .

$f (x) = \frac{x \cdot x^{2} + x (5 x) + 6 x}{3 x^{2} + 9 x}$

Factor $x$ out of $6 x$ .

$f (x) = \frac{x \cdot x^{2} + x (5 x) + x \cdot 6}{3 x^{2} + 9 x}$

Factor $x$ out of $x \cdot x^{2} + x (5 x)$ .

$f (x) = \frac{x (x^{2} + 5 x) + x \cdot 6}{3 x^{2} + 9 x}$

Factor $x$ out of $x (x^{2} + 5 x) + x \cdot 6$ .

$f (x) = \frac{x (x^{2} + 5 x + 6)}{3 x^{2} + 9 x}$

Factor.

Factor $x^{2} + 5 x + 6$ 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 $5$ .

$2, 3$

Write the factored form using these integers.

$f (x) = \frac{x ((x + 2) (x + 3))}{3 x^{2} + 9 x}$

Remove unnecessary parentheses.

$f (x) = \frac{x (x + 2) (x + 3)}{3 x^{2} + 9 x}$

Step 2

Factor $3 x$ out of $3 x^{2} + 9 x$ .

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

$f (x) = \frac{x (x + 2) (x + 3)}{3 x (x) + 9 x}$

Factor $3 x$ out of $9 x$ .

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

Factor $3 x$ out of $3 x (x) + 3 x (3)$ .

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

Step 3

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 5

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

$x, x + 3$

Step 6

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

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{x + 2}{3}$ and simplify.

Substitute $0$ for $x$ to find the $y$ coordinate of the hole.

$\frac{0 + 2}{3}$

Add $0$ and $2$ .

$\frac{2}{3}$

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

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

Simplify.

Add $- 3$ and $2$ .

$\frac{- 1}{3}$

Move the negative in front of the fraction.

$- \frac{1}{3}$

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

$(0, \frac{2}{3}), (- 3, - \frac{1}{3})$

Step 7

Find the Holes in the Graph f(x)=(x^3+5x^2+6x)/(3x^2+9x)

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