Precalculus Examples

$f (x) = x^{3} - 125$

Step 1

Set $x^{3} - 125$ equal to $0$ .

$x^{3} - 125 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Add $125$ to both sides of the equation.

$x^{3} = 125$

Step 2.2

Subtract $125$ from both sides of the equation.

$x^{3} - 125 = 0$

Step 2.3

Factor the left side of the equation.

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Step 2.3.1

Rewrite $125$ as $5^{3}$ .

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

Step 2.3.2

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 $a = x$ and $b = 5$ .

$(x - 5) (x^{2} + x \cdot 5 + 5^{2}) = 0$

Step 2.3.3

Simplify.

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Step 2.3.3.1

Move $5$ to the left of $x$ .

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

Step 2.3.3.2

Raise $5$ to the power of $2$ .

$(x - 5) (x^{2} + 5 x + 25) = 0$

Step 2.4

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

$x - 5 = 0$

$x^{2} + 5 x + 25 = 0$

Step 2.5

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

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Step 2.5.1

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

$x - 5 = 0$

Step 2.5.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.6

Set $x^{2} + 5 x + 25$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x^{2} + 5 x + 25$ equal to $0$ .

$x^{2} + 5 x + 25 = 0$

Step 2.6.2

Solve $x^{2} + 5 x + 25 = 0$ for $x$ .

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Step 2.6.2.1

Use the quadratic formula to find the solutions.

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

Step 2.6.2.2

Substitute the values $a = 1$ , $b = 5$ , and $c = 25$ into the quadratic formula and solve for $x$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot 25)}}{2 \cdot 1}$

Step 2.6.2.3

Simplify.

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Step 2.6.2.3.1

Simplify the numerator.

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Step 2.6.2.3.1.1

Raise $5$ to the power of $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot 25}}{2 \cdot 1}$

Step 2.6.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 25$ .

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Step 2.6.2.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 25}}{2 \cdot 1}$

Step 2.6.2.3.1.2.2

Multiply $- 4$ by $25$ .

$x = \frac{- 5 \pm \sqrt{25 - 100}}{2 \cdot 1}$

Step 2.6.2.3.1.3

Subtract $100$ from $25$ .

$x = \frac{- 5 \pm \sqrt{- 75}}{2 \cdot 1}$

Step 2.6.2.3.1.4

Rewrite $- 75$ as $- 1 (75)$ .

$x = \frac{- 5 \pm \sqrt{- 1 \cdot 75}}{2 \cdot 1}$

Step 2.6.2.3.1.5

Rewrite $\sqrt{- 1 (75)}$ as $\sqrt{- 1} \cdot \sqrt{75}$ .

$x = \frac{- 5 \pm \sqrt{- 1} \cdot \sqrt{75}}{2 \cdot 1}$

Step 2.6.2.3.1.6

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

$x = \frac{- 5 \pm i \cdot \sqrt{75}}{2 \cdot 1}$

Step 2.6.2.3.1.7

Rewrite $75$ as $5^{2} \cdot 3$ .

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Step 2.6.2.3.1.7.1

Factor $25$ out of $75$ .

$x = \frac{- 5 \pm i \cdot \sqrt{25 (3)}}{2 \cdot 1}$

Step 2.6.2.3.1.7.2

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

$x = \frac{- 5 \pm i \cdot \sqrt{5^{2} \cdot 3}}{2 \cdot 1}$

Step 2.6.2.3.1.8

Pull terms out from under the radical.

$x = \frac{- 5 \pm i \cdot (5 \sqrt{3})}{2 \cdot 1}$

Step 2.6.2.3.1.9

Move $5$ to the left of $i$ .

$x = \frac{- 5 \pm 5 i \sqrt{3}}{2 \cdot 1}$

Step 2.6.2.3.2

Multiply $2$ by $1$ .

$x = \frac{- 5 \pm 5 i \sqrt{3}}{2}$

Step 2.6.2.4

The final answer is the combination of both solutions.

$x = - \frac{5 - 5 i \sqrt{3}}{2}, - \frac{5 + 5 i \sqrt{3}}{2}$

Step 2.7

The final solution is all the values that make $(x - 5) (x^{2} + 5 x + 25) = 0$ true.

$x = 5, - \frac{5 - 5 i \sqrt{3}}{2}, - \frac{5 + 5 i \sqrt{3}}{2}$

Step 3