Finite Math Examples

$p^{3} - 25 p = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $p$ out of $p^{3} - 25 p$ .

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Step 1.1.1

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

$p \cdot p^{2} - 25 p = 0$

Step 1.1.2

Factor $p$ out of $- 25 p$ .

$p \cdot p^{2} + p \cdot - 25 = 0$

Step 1.1.3

Factor $p$ out of $p \cdot p^{2} + p \cdot - 25$ .

$p (p^{2} - 25) = 0$

Step 1.2

Rewrite $25$ as $5^{2}$ .

$p (p^{2} - 5^{2}) = 0$

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 $a = p$ and $b = 5$ .

$p ((p + 5) (p - 5)) = 0$

Step 1.3.2

Remove unnecessary parentheses.

$p (p + 5) (p - 5) = 0$

Step 2

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

$p = 0$

$p + 5 = 0$

$p - 5 = 0$

Step 3

Set $p$ equal to $0$ .

$p = 0$

Step 4

Set $p + 5$ equal to $0$ and solve for $p$ .

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Step 4.1

Set $p + 5$ equal to $0$ .

$p + 5 = 0$

Step 4.2

Subtract $5$ from both sides of the equation.

$p = - 5$

Step 5

Set $p - 5$ equal to $0$ and solve for $p$ .

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Step 5.1

Set $p - 5$ equal to $0$ .

$p - 5 = 0$

Step 5.2

Add $5$ to both sides of the equation.

$p = 5$

Step 6

The final solution is all the values that make $p (p + 5) (p - 5) = 0$ true.

$p = 0, - 5, 5$