Pre-Algebra Examples

$w^{4} - w = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $w$ out of $w^{4} - w$ .

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Step 1.1.1

Factor $w$ out of $w^{4}$ .

$w \cdot w^{3} - w = 0$

Step 1.1.2

Factor $w$ out of $- w$ .

$w \cdot w^{3} + w \cdot - 1 = 0$

Step 1.1.3

Factor $w$ out of $w \cdot w^{3} + w \cdot - 1$ .

$w (w^{3} - 1) = 0$

Step 1.2

Rewrite $1$ as $1^{3}$ .

$w (w^{3} - 1^{3}) = 0$

Step 1.3

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = w$ and $b = 1$ .

$w ((w - 1) (w^{2} + w \cdot 1 + 1^{2})) = 0$

Step 1.4

Factor.

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Step 1.4.1

Simplify.

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Step 1.4.1.1

Multiply $w$ by $1$ .

$w ((w - 1) (w^{2} + w + 1^{2})) = 0$

Step 1.4.1.2

One to any power is one.

$w ((w - 1) (w^{2} + w + 1)) = 0$

Step 1.4.2

Remove unnecessary parentheses.

$w (w - 1) (w^{2} + w + 1) = 0$

Step 2

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

$w = 0$

$w - 1 = 0$

$w^{2} + w + 1 = 0$

Step 3

Set $w$ equal to $0$ .

$w = 0$

Step 4

Set $w - 1$ equal to $0$ and solve for $w$ .

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Step 4.1

Set $w - 1$ equal to $0$ .

$w - 1 = 0$

Step 4.2

Add $1$ to both sides of the equation.

$w = 1$

Step 5

Set $w^{2} + w + 1$ equal to $0$ and solve for $w$ .

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Step 5.1

Set $w^{2} + w + 1$ equal to $0$ .

$w^{2} + w + 1 = 0$

Step 5.2

Solve $w^{2} + w + 1 = 0$ for $w$ .

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Step 5.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5.2.2

Substitute the values $a = 1$ , $b = 1$ , and $c = 1$ into the quadratic formula and solve for $w$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 5.2.3

Simplify.

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Step 5.2.3.1

Simplify the numerator.

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Step 5.2.3.1.1

One to any power is one.

$w = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 5.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 5.2.3.1.2.1

Multiply $- 4$ by $1$ .

$w = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 5.2.3.1.2.2

Multiply $- 4$ by $1$ .

$w = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 5.2.3.1.3

Subtract $4$ from $1$ .

$w = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 5.2.3.1.4

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

$w = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 5.2.3.1.5

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

$w = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 5.2.3.1.6

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

$w = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 5.2.3.2

Multiply $2$ by $1$ .

$w = \frac{- 1 \pm i \sqrt{3}}{2}$

Step 5.2.4

The final answer is the combination of both solutions.

$w = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Step 6

The final solution is all the values that make $w (w - 1) (w^{2} + w + 1) = 0$ true.

$w = 0, 1, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$