Trigonometry Examples

$p (x) = 27 x^{3} - 64$

Step 1

Set $27 x^{3} - 64$ equal to $0$ .

$27 x^{3} - 64 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Add $64$ to both sides of the equation.

$27 x^{3} = 64$

Step 2.2

Subtract $64$ from both sides of the equation.

$27 x^{3} - 64 = 0$

Step 2.3

Factor the left side of the equation.

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Step 2.3.1

Rewrite $27 x^{3}$ as ${(3 x)}^{3}$ .

${(3 x)}^{3} - 64 = 0$

Step 2.3.2

Rewrite $64$ as $4^{3}$ .

${(3 x)}^{3} - 4^{3} = 0$

Step 2.3.3

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 $a = 3 x$ and $b = 4$ .

$(3 x - 4) ({(3 x)}^{2} + 3 x \cdot 4 + 4^{2}) = 0$

Step 2.3.4

Simplify.

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Step 2.3.4.1

Apply the product rule to $3 x$ .

$(3 x - 4) (3^{2} x^{2} + 3 x \cdot 4 + 4^{2}) = 0$

Step 2.3.4.2

Raise $3$ to the power of $2$ .

$(3 x - 4) (9 x^{2} + 3 x \cdot 4 + 4^{2}) = 0$

Step 2.3.4.3

Multiply $4$ by $3$ .

$(3 x - 4) (9 x^{2} + 12 x + 4^{2}) = 0$

Step 2.3.4.4

Raise $4$ to the power of $2$ .

$(3 x - 4) (9 x^{2} + 12 x + 16) = 0$

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 $0$ .

$3 x - 4 = 0$

$9 x^{2} + 12 x + 16 = 0$

Step 2.5

Set $3 x - 4$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $3 x - 4$ equal to $0$ .

$3 x - 4 = 0$

Step 2.5.2

Solve $3 x - 4 = 0$ for $x$ .

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Step 2.5.2.1

Add $4$ to both sides of the equation.

$3 x = 4$

Step 2.5.2.2

Divide each term in $3 x = 4$ by $3$ and simplify.

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Step 2.5.2.2.1

Divide each term in $3 x = 4$ by $3$ .

$\frac{3 x}{3} = \frac{4}{3}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $3$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{4}{3}$

Step 2.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{4}{3}$

Step 2.6

Set $9 x^{2} + 12 x + 16$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $9 x^{2} + 12 x + 16$ equal to $0$ .

$9 x^{2} + 12 x + 16 = 0$

Step 2.6.2

Solve $9 x^{2} + 12 x + 16 = 0$ 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 $a = 9$ , $b = 12$ , and $c = 16$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (9 \cdot 16)}}{2 \cdot 9}$

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 $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 9 \cdot 16}}{2 \cdot 9}$

Step 2.6.2.3.1.2

Multiply $- 4 \cdot 9 \cdot 16$ .

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Step 2.6.2.3.1.2.1

Multiply $- 4$ by $9$ .

$x = \frac{- 12 \pm \sqrt{144 - 36 \cdot 16}}{2 \cdot 9}$

Step 2.6.2.3.1.2.2

Multiply $- 36$ by $16$ .

$x = \frac{- 12 \pm \sqrt{144 - 576}}{2 \cdot 9}$

Step 2.6.2.3.1.3

Subtract $576$ from $144$ .

$x = \frac{- 12 \pm \sqrt{- 432}}{2 \cdot 9}$

Step 2.6.2.3.1.4

Rewrite $- 432$ as $- 1 (432)$ .

$x = \frac{- 12 \pm \sqrt{- 1 \cdot 432}}{2 \cdot 9}$

Step 2.6.2.3.1.5

Rewrite $\sqrt{- 1 (432)}$ as $\sqrt{- 1} \cdot \sqrt{432}$ .

$x = \frac{- 12 \pm \sqrt{- 1} \cdot \sqrt{432}}{2 \cdot 9}$

Step 2.6.2.3.1.6

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

$x = \frac{- 12 \pm i \cdot \sqrt{432}}{2 \cdot 9}$

Step 2.6.2.3.1.7

Rewrite $432$ as $12^{2} \cdot 3$ .

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Step 2.6.2.3.1.7.1

Factor $144$ out of $432$ .

$x = \frac{- 12 \pm i \cdot \sqrt{144 (3)}}{2 \cdot 9}$

Step 2.6.2.3.1.7.2

Rewrite $144$ as $12^{2}$ .

$x = \frac{- 12 \pm i \cdot \sqrt{12^{2} \cdot 3}}{2 \cdot 9}$

Step 2.6.2.3.1.8

Pull terms out from under the radical.

$x = \frac{- 12 \pm i \cdot (12 \sqrt{3})}{2 \cdot 9}$

Step 2.6.2.3.1.9

Move $12$ to the left of $i$ .

$x = \frac{- 12 \pm 12 i \sqrt{3}}{2 \cdot 9}$

Step 2.6.2.3.2

Multiply $2$ by $9$ .

$x = \frac{- 12 \pm 12 i \sqrt{3}}{18}$

Step 2.6.2.3.3

Simplify $\frac{- 12 \pm 12 i \sqrt{3}}{18}$ .

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

Step 2.6.2.4

The final answer is the combination of both solutions.

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

Step 2.7

The final solution is all the values that make $(3 x - 4) (9 x^{2} + 12 x + 16) = 0$ true.

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

Step 3