Algebra Examples

$x^{5} = x^{3} + 12 x$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $x^{3}$ from both sides of the equation.

$x^{5} - x^{3} = 12 x$

Step 1.2

Subtract $12 x$ from both sides of the equation.

$x^{5} - x^{3} - 12 x = 0$

Step 2

Factor $x$ out of $x^{5} - x^{3} - 12 x$ .

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Step 2.1

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

$x \cdot x^{4} - x^{3} - 12 x = 0$

Step 2.2

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

$x \cdot x^{4} + x (- x^{2}) - 12 x = 0$

Step 2.3

Factor $x$ out of $- 12 x$ .

$x \cdot x^{4} + x (- x^{2}) + x \cdot - 12 = 0$

Step 2.4

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

$x (x^{4} - x^{2}) + x \cdot - 12 = 0$

Step 2.5

Factor $x$ out of $x (x^{4} - x^{2}) + x \cdot - 12$ .

$x (x^{4} - x^{2} - 12) = 0$

Step 3

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

$x ({(x^{2})}^{2} - x^{2} - 12) = 0$

Step 4

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$x (u^{2} - u - 12) = 0$

Step 5

Factor $u^{2} - u - 12$ using the AC method.

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Step 5.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Step 5.2

Write the factored form using these integers.

$x ((u - 4) (u + 3)) = 0$

Step 6

Replace all occurrences of $u$ with $x^{2}$ .

$x ((x^{2} - 4) (x^{2} + 3)) = 0$

Step 7

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

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

Step 8

Factor.

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Step 8.1

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 $b = 2$ .

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

Step 8.2

Remove unnecessary parentheses.

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

Step 9

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

$x = 0$

$x + 2 = 0$

$x - 2 = 0$

$x^{2} + 3 = 0$

Step 10

Set $x$ equal to $0$ .

$x = 0$

Step 11

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 11.1

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

$x + 2 = 0$

Step 11.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 12

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 12.1

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

$x - 2 = 0$

Step 12.2

Add $2$ to both sides of the equation.

$x = 2$

Step 13

Set $x^{2} + 3$ equal to $0$ and solve for $x$ .

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Step 13.1

Set $x^{2} + 3$ equal to $0$ .

$x^{2} + 3 = 0$

Step 13.2

Solve $x^{2} + 3 = 0$ for $x$ .

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Step 13.2.1

Subtract $3$ from both sides of the equation.

$x^{2} = - 3$

Step 13.2.2

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

$x = \pm \sqrt{- 3}$

Step 13.2.3

Simplify $\pm \sqrt{- 3}$ .

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Step 13.2.3.1

Rewrite $- 3$ as $- 1 (3)$ .

$x = \pm \sqrt{- 1 (3)}$

Step 13.2.3.2

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{3}$

Step 13.2.3.3

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

$x = \pm i \sqrt{3}$

Step 13.2.4

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

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Step 13.2.4.1

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

$x = i \sqrt{3}$

Step 13.2.4.2

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

$x = - i \sqrt{3}$

Step 13.2.4.3

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

$x = i \sqrt{3}, - i \sqrt{3}$

Step 14

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

$x = 0, - 2, 2, i \sqrt{3}, - i \sqrt{3}$