Calculus Examples

$4 x^{3} - 16 x = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $4 x$ out of $4 x^{3} - 16 x$ .

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Step 1.1.1

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

$4 x (x^{2}) - 16 x = 0$

Step 1.1.2

Factor $4 x$ out of $- 16 x$ .

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

Step 1.1.3

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

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

Step 1.2

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

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

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 $b = 2$ .

$4 x ((x + 2) (x - 2)) = 0$

Step 1.3.2

Remove unnecessary parentheses.

$4 x (x + 2) (x - 2) = 0$

Step 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 3

Set $x$ equal to $0$ .

$x = 0$

Step 4

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

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Step 4.1

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

$x + 2 = 0$

Step 4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5

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

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Step 5.1

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

$x - 2 = 0$

Step 5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 6

The final solution is all the values that make $4 x (x + 2) (x - 2) = 0$ true.

$x = 0, - 2, 2$