Algebra Examples

$x^{3} - 16 x = 0$

Step 1

Factor $x$ out of $x^{3} - 16 x$ .

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Step 1.1

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

$x \cdot x^{2} - 16 x = 0$

Step 1.2

Factor $x$ out of $- 16 x$ .

$x \cdot x^{2} + x \cdot - 16 = 0$

Step 1.3

Factor $x$ out of $x \cdot x^{2} + x \cdot - 16$ .

$x (x^{2} - 16) = 0$

Step 2

Rewrite $16$ as $4^{2}$ .

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

Step 3

Factor.

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Step 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 $b = 4$ .

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

Step 3.2

Remove unnecessary parentheses.

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

Step 4

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

$x = 0$

$x + 4 = 0$

$x - 4 = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 6.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 7

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 7.2

Add $4$ to both sides of the equation.

$x = 4$

Step 8

The final solution is all the values that make $x (x + 4) (x - 4) = 0$ true.

$x = 0, - 4, 4$