Algebra Examples

$\frac{x^{4} + 8 x^{2} + 7}{3 x^{5} - 3 x}$

Step 1

Set the denominator in $\frac{x^{4} + 8 x^{2} + 7}{3 x^{5} - 3 x}$ equal to $0$ to find where the expression is undefined.

$3 x^{5} - 3 x = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Factor $3 x$ out of $3 x^{5} - 3 x$ .

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Step 2.1.1.1

Factor $3 x$ out of $3 x^{5}$ .

$3 x (x^{4}) - 3 x = 0$

Step 2.1.1.2

Factor $3 x$ out of $- 3 x$ .

$3 x (x^{4}) + 3 x (- 1) = 0$

Step 2.1.1.3

Factor $3 x$ out of $3 x (x^{4}) + 3 x (- 1)$ .

$3 x (x^{4} - 1) = 0$

Step 2.1.2

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$3 x ({(x^{2})}^{2} - 1) = 0$

Step 2.1.3

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

$3 x ({(x^{2})}^{2} - 1^{2}) = 0$

Step 2.1.4

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

$3 x ((x^{2} + 1) (x^{2} - 1)) = 0$

Step 2.1.5

Factor.

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Step 2.1.5.1

Simplify.

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Step 2.1.5.1.1

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

$3 x ((x^{2} + 1) (x^{2} - 1^{2})) = 0$

Step 2.1.5.1.2

Factor.

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Step 2.1.5.1.2.1

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 $b = 1$ .

$3 x ((x^{2} + 1) ((x + 1) (x - 1))) = 0$

Step 2.1.5.1.2.2

Remove unnecessary parentheses.

$3 x ((x^{2} + 1) (x + 1) (x - 1)) = 0$

Step 2.1.5.2

Remove unnecessary parentheses.

$3 x (x^{2} + 1) (x + 1) (x - 1) = 0$

Step 2.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x^{2} + 1 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Step 2.4.2

Solve $x^{2} + 1 = 0$ for $x$ .

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Step 2.4.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 2.4.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 1}$

Step 2.4.2.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 2.4.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.4.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = i$

Step 2.4.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i$

Step 2.4.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = i, - i$

Step 2.5

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 2.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.6

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 2.6.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.7

The final solution is all the values that make $3 x (x^{2} + 1) (x + 1) (x - 1) = 0$ true.

$x = 0, i, - i, - 1, 1$

Step 3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 1, x = 0, x = 1$

Step 4