Find the Holes in the Graph f(x)=((x-3)(x-6)(x+4))/((x-3)^2(x-6)(x-8))

$f (x) = \frac{(x - 3) (x - 6) (x + 4)}{{(x - 3)}^{2} (x - 6) (x - 8)}$

Step 1

Cancel the common factor of $x - 3$ and ${(x - 3)}^{2}$ .

Factor $x - 3$ out of $(x - 3) (x - 6) (x + 4)$ .

$f (x) = \frac{(x - 3) ((x - 6) (x + 4))}{{(x - 3)}^{2} (x - 6) (x - 8)}$

Cancel the common factors.

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

$f (x) = \frac{(x - 3) ((x - 6) (x + 4))}{(x - 3) (((x - 3) (x - 6)) (x - 8))}$

Cancel the common factor.

$f (x) = \frac{(x - 3) ((x - 6) (x + 4))}{(x - 3) (((x - 3) (x - 6)) (x - 8))}$

Rewrite the expression.

$f (x) = \frac{(x - 6) (x + 4)}{((x - 3) (x - 6)) (x - 8)}$

Step 2

Cancel the common factor of $x - 6$ .

Cancel the common factor.

$f (x) = \frac{(x - 6) (x + 4)}{(x - 3) (x - 6) (x - 8)}$

Rewrite the expression.

$f (x) = \frac{x + 4}{(x - 3) (x - 8)}$

Step 3

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

$x - 6$

Step 4

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 4}{(x - 3) (x - 8)}$ .

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

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Substitute $6$ for $x$ in $\frac{x + 4}{(x - 3) (x - 8)}$ and simplify.

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

$\frac{6 + 4}{(6 - 3) (6 - 8)}$

Simplify.

Cancel the common factor of $6 + 4$ and $6 - 8$ .

Factor $2$ out of $6$ .

$\frac{2 (3) + 4}{(6 - 3) (6 - 8)}$

Factor $2$ out of $4$ .

$\frac{2 \cdot 3 + 2 \cdot 2}{(6 - 3) (6 - 8)}$

Factor $2$ out of $2 \cdot 3 + 2 \cdot 2$ .

$\frac{2 \cdot (3 + 2)}{(6 - 3) (6 - 8)}$

Cancel the common factors.

Factor $2$ out of $(6 - 3) (6 - 8)$ .

$\frac{2 \cdot (3 + 2)}{2 ((6 - 3) (3 - 4))}$

Cancel the common factor.

$\frac{2 \cdot (3 + 2)}{2 ((6 - 3) (3 - 4))}$

Rewrite the expression.

$\frac{3 + 2}{(6 - 3) (3 - 4)}$

Add $3$ and $2$ .

$\frac{5}{(6 - 3) (3 - 4)}$

Simplify the denominator.

Subtract $3$ from $6$ .

$\frac{5}{3 (3 - 4)}$

Subtract $4$ from $3$ .

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

Multiply $3$ by $- 1$ .

$\frac{5}{- 3}$

Move the negative in front of the fraction.

$- \frac{5}{3}$

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

$(6, - \frac{5}{3})$

Step 5

Find the Holes in the Graph f(x)=((x-3)(x-6)(x+4))/((x-3)^2(x-6)(x-8))

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