Precalculus Examples

$f (x) = x^{5} - 16 x$

Step 1

Set $x^{5} - 16 x$ equal to $0$ .

$x^{5} - 16 x = 0$

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 $x^{5} - 16 x$ .

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Step 2.1.1.1

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

$x \cdot x^{4} - 16 x = 0$

Step 2.1.1.2

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

$x \cdot x^{4} + x \cdot - 16 = 0$

Step 2.1.1.3

Factor $x$ out of $x \cdot x^{4} + x \cdot - 16$ .

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

Step 2.1.2

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

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

Step 2.1.3

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

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

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 $b = 4$ .

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

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

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

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 $b = 2$ .

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

Step 2.1.5.1.2.2

Remove unnecessary parentheses.

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

Step 2.1.5.2

Remove unnecessary parentheses.

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

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 $0$ .

$x = 0$

$x^{2} + 4 = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

Set $x^{2} + 4$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x^{2} + 4$ equal to $0$ .

$x^{2} + 4 = 0$

Step 2.4.2

Solve $x^{2} + 4 = 0$ for $x$ .

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Step 2.4.2.1

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

Step 2.4.2.2

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

$x = \pm \sqrt{- 4}$

Step 2.4.2.3

Simplify $\pm \sqrt{- 4}$ .

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Step 2.4.2.3.1

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Step 2.4.2.3.2

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{4}$

Step 2.4.2.3.3

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

$x = \pm i \cdot \sqrt{4}$

Step 2.4.2.3.4

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

$x = \pm i \cdot \sqrt{2^{2}}$

Step 2.4.2.3.5

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 2$

Step 2.4.2.3.6

Move $2$ to the left of $i$ .

$x = \pm 2 i$

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.

$x = 2 i$

Step 2.4.2.4.2

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

$x = - 2 i$

Step 2.4.2.4.3

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

$x = 2 i, - 2 i$

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

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.6

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

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Step 2.6.1

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

$x - 2 = 0$

Step 2.6.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.7

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

$x = 0, 2 i, - 2 i, - 2, 2$

Step 3