Precalculus Examples

$8 z^{3} - 125 = 0$

Step 1

Add $125$ to both sides of the equation.

$8 z^{3} = 125$

Step 2

Subtract $125$ from both sides of the equation.

$8 z^{3} - 125 = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Rewrite $8 z^{3}$ as ${(2 z)}^{3}$ .

${(2 z)}^{3} - 125 = 0$

Step 3.2

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

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

Step 3.3

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 $a = 2 z$ and $b = 5$ .

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

Step 3.4

Simplify.

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Step 3.4.1

Apply the product rule to $2 z$ .

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

Step 3.4.2

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

$(2 z - 5) (4 z^{2} + 2 z \cdot 5 + 5^{2}) = 0$

Step 3.4.3

Multiply $5$ by $2$ .

$(2 z - 5) (4 z^{2} + 10 z + 5^{2}) = 0$

Step 3.4.4

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

$(2 z - 5) (4 z^{2} + 10 z + 25) = 0$

Step 4

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

$2 z - 5 = 0$

$4 z^{2} + 10 z + 25 = 0$

Step 5

Set $2 z - 5$ equal to $0$ and solve for $z$ .

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Step 5.1

Set $2 z - 5$ equal to $0$ .

$2 z - 5 = 0$

Step 5.2

Solve $2 z - 5 = 0$ for $z$ .

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Step 5.2.1

Add $5$ to both sides of the equation.

$2 z = 5$

Step 5.2.2

Divide each term in $2 z = 5$ by $2$ and simplify.

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Step 5.2.2.1

Divide each term in $2 z = 5$ by $2$ .

$\frac{2 z}{2} = \frac{5}{2}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $2$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{2 z}{2} = \frac{5}{2}$

Step 5.2.2.2.1.2

Divide $z$ by $1$ .

$z = \frac{5}{2}$

Step 6

Set $4 z^{2} + 10 z + 25$ equal to $0$ and solve for $z$ .

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Step 6.1

Set $4 z^{2} + 10 z + 25$ equal to $0$ .

$4 z^{2} + 10 z + 25 = 0$

Step 6.2

Solve $4 z^{2} + 10 z + 25 = 0$ for $z$ .

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Step 6.2.1

Use the quadratic formula to find the solutions.

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

Step 6.2.2

Substitute the values $a = 4$ , $b = 10$ , and $c = 25$ into the quadratic formula and solve for $z$ .

$\frac{- 10 \pm \sqrt{10^{2} - 4 \cdot (4 \cdot 25)}}{2 \cdot 4}$

Step 6.2.3

Simplify.

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Step 6.2.3.1

Simplify the numerator.

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Step 6.2.3.1.1

Raise $10$ to the power of $2$ .

$z = \frac{- 10 \pm \sqrt{100 - 4 \cdot 4 \cdot 25}}{2 \cdot 4}$

Step 6.2.3.1.2

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

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Step 6.2.3.1.2.1

Multiply $- 4$ by $4$ .

$z = \frac{- 10 \pm \sqrt{100 - 16 \cdot 25}}{2 \cdot 4}$

Step 6.2.3.1.2.2

Multiply $- 16$ by $25$ .

$z = \frac{- 10 \pm \sqrt{100 - 400}}{2 \cdot 4}$

Step 6.2.3.1.3

Subtract $400$ from $100$ .

$z = \frac{- 10 \pm \sqrt{- 300}}{2 \cdot 4}$

Step 6.2.3.1.4

Rewrite $- 300$ as $- 1 (300)$ .

$z = \frac{- 10 \pm \sqrt{- 1 \cdot 300}}{2 \cdot 4}$

Step 6.2.3.1.5

Rewrite $\sqrt{- 1 (300)}$ as $\sqrt{- 1} \cdot \sqrt{300}$ .

$z = \frac{- 10 \pm \sqrt{- 1} \cdot \sqrt{300}}{2 \cdot 4}$

Step 6.2.3.1.6

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

$z = \frac{- 10 \pm i \cdot \sqrt{300}}{2 \cdot 4}$

Step 6.2.3.1.7

Rewrite $300$ as $10^{2} \cdot 3$ .

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Step 6.2.3.1.7.1

Factor $100$ out of $300$ .

$z = \frac{- 10 \pm i \cdot \sqrt{100 (3)}}{2 \cdot 4}$

Step 6.2.3.1.7.2

Rewrite $100$ as $10^{2}$ .

$z = \frac{- 10 \pm i \cdot \sqrt{10^{2} \cdot 3}}{2 \cdot 4}$

Step 6.2.3.1.8

Pull terms out from under the radical.

$z = \frac{- 10 \pm i \cdot (10 \sqrt{3})}{2 \cdot 4}$

Step 6.2.3.1.9

Move $10$ to the left of $i$ .

$z = \frac{- 10 \pm 10 i \sqrt{3}}{2 \cdot 4}$

Step 6.2.3.2

Multiply $2$ by $4$ .

$z = \frac{- 10 \pm 10 i \sqrt{3}}{8}$

Step 6.2.3.3

Simplify $\frac{- 10 \pm 10 i \sqrt{3}}{8}$ .

$z = \frac{- 5 \pm 5 i \sqrt{3}}{4}$

Step 6.2.4

The final answer is the combination of both solutions.

$z = - \frac{5 - 5 i \sqrt{3}}{4}, - \frac{5 + 5 i \sqrt{3}}{4}$

Step 7

The final solution is all the values that make $(2 z - 5) (4 z^{2} + 10 z + 25) = 0$ true.

$z = \frac{5}{2}, - \frac{5 - 5 i \sqrt{3}}{4}, - \frac{5 + 5 i \sqrt{3}}{4}$