Basic Math Examples

$6 y^{3} - 24 y = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $6 y$ out of $6 y^{3} - 24 y$ .

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Step 1.1.1

Factor $6 y$ out of $6 y^{3}$ .

$6 y (y^{2}) - 24 y = 0$

Step 1.1.2

Factor $6 y$ out of $- 24 y$ .

$6 y (y^{2}) + 6 y (- 4) = 0$

Step 1.1.3

Factor $6 y$ out of $6 y (y^{2}) + 6 y (- 4)$ .

$6 y (y^{2} - 4) = 0$

Step 1.2

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

$6 y (y^{2} - 2^{2}) = 0$

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 $a = y$ and $b = 2$ .

$6 y ((y + 2) (y - 2)) = 0$

Step 1.3.2

Remove unnecessary parentheses.

$6 y (y + 2) (y - 2) = 0$

Step 2

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

$y = 0$

$y + 2 = 0$

$y - 2 = 0$

Step 3

Set $y$ equal to $0$ .

$y = 0$

Step 4

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

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Step 4.1

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

$y + 2 = 0$

Step 4.2

Subtract $2$ from both sides of the equation.

$y = - 2$

Step 5

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

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Step 5.1

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

$y - 2 = 0$

Step 5.2

Add $2$ to both sides of the equation.

$y = 2$

Step 6

The final solution is all the values that make $6 y (y + 2) (y - 2) = 0$ true.

$y = 0, - 2, 2$