Algebra Examples

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

Step 1

Factor $x^{3} - 4 x$ .

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Step 1.1

Factor $x$ out of $x^{3} - 4 x$ .

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Step 1.1.1

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

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

Step 1.1.2

Factor $x$ out of $- 4 x$ .

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

Step 1.1.3

Factor $x$ out of $x \cdot x^{2} + x \cdot - 4$ .

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

Step 1.2

Rewrite $4$ as $2^{2}$ .

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

Step 1.3

Factor.

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Step 1.3.1

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 $b = 2$ .

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

Step 1.3.2

Remove unnecessary parentheses.

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

Step 2

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

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Step 2.1

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

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

Step 2.2

Factor $3 x$ out of $6 x$ .

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

Step 2.3

Factor $3 x$ out of $3 x (- x) + 3 x (2)$ .

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

Step 3

Cancel the common factor of $x$ .

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Step 3.1

Cancel the common factor.

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

Step 3.2

Rewrite the expression.

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

Step 4

Cancel the common factor of $x - 2$ and $- x + 2$ .

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Step 4.1

Factor $- 1$ out of $x$ .

$f (x) = \frac{(x + 2) (- 1 (- x) - 2)}{3 (- x + 2)}$

Step 4.2

Rewrite $- 2$ as $- 1 (2)$ .

$f (x) = \frac{(x + 2) (- 1 (- x) - 1 \cdot 2)}{3 (- x + 2)}$

Step 4.3

Factor $- 1$ out of $- 1 (- x) - 1 (2)$ .

$f (x) = \frac{(x + 2) (- 1 (- x + 2))}{3 (- x + 2)}$

Step 4.4

Cancel the common factor.

$f (x) = \frac{(x + 2) (- 1 (- x + 2))}{3 (- x + 2)}$

Step 4.5

Rewrite the expression.

$f (x) = \frac{(x + 2) \cdot (- 1)}{3}$

Step 5

Multiply $x + 2$ by $- 1$ .

$f (x) = \frac{(x + 2) \cdot - 1}{3}$

Step 6

Move $- 1$ to the left of $x + 2$ .

$f (x) = \frac{- 1 \cdot (x + 2)}{3}$

Step 7

Move the negative in front of the fraction.

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

Step 8

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

$x, - x + 2$

Step 9

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

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Step 9.1

Set $x$ equal to $0$ .

$x = 0$

Step 9.2

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

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Step 9.2.1

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

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

Step 9.2.2

Add $0$ and $2$ .

$- \frac{2}{3}$

Step 9.3

Set $- x + 2$ equal to $0$ .

$- x + 2 = 0$

Step 9.4

Solve $- x + 2 = 0$ for $x$ .

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Step 9.4.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 9.4.2

Divide each term in $- x = - 2$ by $- 1$ and simplify.

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Step 9.4.2.1

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Step 9.4.2.2

Simplify the left side.

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Step 9.4.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Step 9.4.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Step 9.4.2.3

Simplify the right side.

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Step 9.4.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 9.5

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

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Step 9.5.1

Substitute $2$ for $x$ to find the $y$ coordinate of the hole.

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

Step 9.5.2

Add $2$ and $2$ .

$- \frac{4}{3}$

Step 9.6

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

$(0, - \frac{2}{3}), (2, - \frac{4}{3})$

Step 10