Algebra Examples

$y = x^{3} - 4 x$

Step 1

Set $x^{3} - 4 x$ equal to $0$ .

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

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

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Step 2.1.1.1

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

$x \cdot x^{2} - 4 x = 0$

Step 2.1.1.2

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

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

Step 2.1.1.3

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

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

Step 2.1.2

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

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

Step 2.1.3

Factor.

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Step 2.1.3.1

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 $b = 2$ .

$x ((x + 2) (x - 2)) = 0$

Step 2.1.3.2

Remove unnecessary parentheses.

$x (x + 2) (x - 2) = 0$

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 $0$ .

$x = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

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

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Step 2.4.1

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

$x + 2 = 0$

Step 2.4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.5

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

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Step 2.5.1

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

$x - 2 = 0$

Step 2.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.6

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

$x = 0, - 2, 2$

Step 3

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