Find the Holes in the Graph f(x)=(x^3-6x^2+8x)/(x^2-2x-8)

$f (x) = \frac{x^{3} - 6 x^{2} + 8 x}{x^{2} - 2 x - 8}$

Factor $x^{3} - 6 x^{2} + 8 x$ .

Factor $x$ out of $x^{3} - 6 x^{2} + 8 x$ .

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

$f (x) = \frac{x \cdot x^{2} - 6 x^{2} + 8 x}{x^{2} - 2 x - 8}$

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

$f (x) = \frac{x \cdot x^{2} + x (- 6 x) + 8 x}{x^{2} - 2 x - 8}$

Factor $x$ out of $8 x$ .

$f (x) = \frac{x \cdot x^{2} + x (- 6 x) + x \cdot 8}{x^{2} - 2 x - 8}$

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

$f (x) = \frac{x (x^{2} - 6 x) + x \cdot 8}{x^{2} - 2 x - 8}$

Factor $x$ out of $x (x^{2} - 6 x) + x \cdot 8$ .

$f (x) = \frac{x (x^{2} - 6 x + 8)}{x^{2} - 2 x - 8}$

Factor.

Factor $x^{2} - 6 x + 8$ 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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Write the factored form using these integers.

$f (x) = \frac{x ((x - 4) (x - 2))}{x^{2} - 2 x - 8}$

Remove unnecessary parentheses.

$f (x) = \frac{x (x - 4) (x - 2)}{x^{2} - 2 x - 8}$

Factor $x^{2} - 2 x - 8$ 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 $- 8$ and whose sum is $- 2$ .

$- 4, 2$

Write the factored form using these integers.

$f (x) = \frac{x (x - 4) (x - 2)}{(x - 4) (x + 2)}$

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$f (x) = \frac{x (x - 4) (x - 2)}{(x - 4) (x + 2)}$

Rewrite the expression.

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

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

$x - 4$

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

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Substitute $4$ for $x$ in $\frac{x (x - 2)}{x + 2}$ and simplify.

Substitute $4$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Cancel the common factor of $4$ and $4 + 2$ .

Factor $2$ out of $4 (4 - 2)$ .

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

Cancel the common factors.

Factor $2$ out of $4$ .

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

Factor $2$ out of $2$ .

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Subtract $2$ from $4$ .

$\frac{2 \cdot 2}{2 + 1}$

Add $2$ and $1$ .

$\frac{2 \cdot 2}{3}$

Multiply $2$ by $2$ .

$\frac{4}{3}$

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

$(4, \frac{4}{3})$

Find the Holes in the Graph f(x)=(x^3-6x^2+8x)/(x^2-2x-8)

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