Precalculus Examples

$x^{5} = 25 x^{3}$

Step 1

Subtract $25 x^{3}$ from both sides of the equation.

$x^{5} - 25 x^{3} = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $x^{3}$ out of $x^{5} - 25 x^{3}$ .

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Step 2.1.1

Factor $x^{3}$ out of $x^{5}$ .

$x^{3} x^{2} - 25 x^{3} = 0$

Step 2.1.2

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

$x^{3} x^{2} + x^{3} \cdot - 25 = 0$

Step 2.1.3

Factor $x^{3}$ out of $x^{3} x^{2} + x^{3} \cdot - 25$ .

$x^{3} (x^{2} - 25) = 0$

Step 2.2

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

$x^{3} (x^{2} - 5^{2}) = 0$

Step 2.3

Factor.

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Step 2.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

$x^{3} ((x + 5) (x - 5)) = 0$

Step 2.3.2

Remove unnecessary parentheses.

$x^{3} (x + 5) (x - 5) = 0$

Step 3

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

$x^{3} = 0$

$x + 5 = 0$

$x - 5 = 0$

Step 4

Set $x^{3}$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x^{3}$ equal to $0$ .

$x^{3} = 0$

Step 4.2

Solve $x^{3} = 0$ for $x$ .

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Step 4.2.1

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

$x = \sqrt[3]{0}$

Step 4.2.2

Simplify $\sqrt[3]{0}$ .

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Step 4.2.2.1

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Step 4.2.2.2

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

$x = 0$

Step 5

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

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Step 5.1

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

$x + 5 = 0$

Step 5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6

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

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Step 6.1

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

$x - 5 = 0$

Step 6.2

Add $5$ to both sides of the equation.

$x = 5$

Step 7

The final solution is all the values that make $x^{3} (x + 5) (x - 5) = 0$ true.

$x = 0, - 5, 5$