Find the Holes in the Graph f(x)=(x^3-7x^2-x+7)/(x-7)

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

Step 1

Factor $x^{3} - 7 x^{2} - x + 7$ .

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $x - 7$ .

$f (x) = \frac{(x - 7) (x^{2} - 1)}{x - 7}$

Rewrite $1$ as $1^{2}$ .

$f (x) = \frac{(x - 7) (x^{2} - 1^{2})}{x - 7}$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$f (x) = \frac{(x - 7) ((x + 1) (x - 1))}{x - 7}$

Remove unnecessary parentheses.

$f (x) = \frac{(x - 7) (x + 1) (x - 1)}{x - 7}$

Step 2

Cancel the common factor of $x - 7$ .

Cancel the common factor.

$f (x) = \frac{(x - 7) (x + 1) (x - 1)}{x - 7}$

Divide $(x + 1) (x - 1)$ by $1$ .

$f (x) = (x + 1) (x - 1)$

Step 3

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

$x - 7$

Step 4

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $(x + 1) (x - 1)$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Substitute $7$ for $x$ in $(x + 1) (x - 1)$ and simplify.

Substitute $7$ for $x$ to find the $y$ coordinate of the hole.

$(7 + 1) (7 - 1)$

Simplify.

Add $7$ and $1$ .

$8 (7 - 1)$

Subtract $1$ from $7$ .

$8 \cdot 6$

Multiply $8$ by $6$ .

$48$

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

$(7, 48)$

Step 5