Precalculus Examples

$w^{3} - 9 w = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $w$ out of $w^{3} - 9 w$ .

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Step 1.1.1

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

$w \cdot w^{2} - 9 w = 0$

Step 1.1.2

Factor $w$ out of $- 9 w$ .

$w \cdot w^{2} + w \cdot - 9 = 0$

Step 1.1.3

Factor $w$ out of $w \cdot w^{2} + w \cdot - 9$ .

$w (w^{2} - 9) = 0$

Step 1.2

Rewrite $9$ as $3^{2}$ .

$w (w^{2} - 3^{2}) = 0$

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 $a = w$ and $b = 3$ .

$w ((w + 3) (w - 3)) = 0$

Step 1.3.2

Remove unnecessary parentheses.

$w (w + 3) (w - 3) = 0$

Step 2

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

$w = 0$

$w + 3 = 0$

$w - 3 = 0$

Step 3

Set $w$ equal to $0$ .

$w = 0$

Step 4

Set $w + 3$ equal to $0$ and solve for $w$ .

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Step 4.1

Set $w + 3$ equal to $0$ .

$w + 3 = 0$

Step 4.2

Subtract $3$ from both sides of the equation.

$w = - 3$

Step 5

Set $w - 3$ equal to $0$ and solve for $w$ .

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Step 5.1

Set $w - 3$ equal to $0$ .

$w - 3 = 0$

Step 5.2

Add $3$ to both sides of the equation.

$w = 3$

Step 6

The final solution is all the values that make $w (w + 3) (w - 3) = 0$ true.

$w = 0, - 3, 3$