Algebra Examples

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

Step 1

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

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Step 1.1

Factor out the greatest common factor from each group.

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Step 1.1.1

Group the first two terms and the last two terms.

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

Step 1.1.2

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

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

Step 1.2

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

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

Step 1.3

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

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

Step 1.4

Factor.

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Step 1.4.1

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 $b = 1$ .

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

Step 1.4.2

Remove unnecessary parentheses.

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

Step 2

Cancel the common factor of $x - 7$ .

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Step 2.1

Cancel the common factor.

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

Step 2.2

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

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Step 4.1

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

$x - 7 = 0$

Step 4.2

Add $7$ to both sides of the equation.

$x = 7$

Step 4.3

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

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Step 4.3.1

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

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

Step 4.3.2

Simplify.

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Step 4.3.2.1

Add $7$ and $1$ .

$8 (7 - 1)$

Step 4.3.2.2

Subtract $1$ from $7$ .

$8 \cdot 6$

Step 4.3.2.3

Multiply $8$ by $6$ .

$48$

Step 4.4

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

$(7, 48)$

Step 5