Find the Roots (Zeros) f(x)=x^3+125

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

Step 1

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

$x^{3} + 125 = 0$

Step 2

Solve for $x$ .

Subtract $125$ from both sides of the equation.

$x^{3} = - 125$

Add $125$ to both sides of the equation.

$x^{3} + 125 = 0$

Factor the left side of the equation.

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

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

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 5$ .

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

Simplify.

Multiply $5$ by $- 1$ .

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

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

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

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

$x + 5 = 0$

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

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

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

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

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $25$ .

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

Subtract $100$ from $25$ .

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

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

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

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

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

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

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

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

Factor $25$ out of $75$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

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

The final answer is the combination of both solutions.

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

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

Find the Roots (Zeros) f(x)=x^3+125

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