Solve Using the Quadratic Formula 8x^3-125=0

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

Add $125$ to both sides of the equation.

$8 x^{3} = 125$

Subtract $125$ from both sides of the equation.

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

Factor the left side of the equation.

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

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

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

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

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

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

Simplify.

Apply the product rule to $2 x$ .

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

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

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

Multiply $5$ by $2$ .

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

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

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

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

$2 x - 5 = 0$

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

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

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

$2 x - 5 = 0$

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

Add $5$ to both sides of the equation.

$2 x = 5$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

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

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $4$ .

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

Multiply $- 16$ by $25$ .

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

Subtract $400$ from $100$ .

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

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

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

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

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

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

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

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

Factor $100$ out of $300$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $4$ .

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

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

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

The final answer is the combination of both solutions.

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

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

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

Solve Using the Quadratic Formula 8x^3-125=0

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