Precalculus Examples

$x^{6} - 13 x^{3} + 40 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Rewrite $x^{6}$ as ${(x^{3})}^{2}$ .

${(x^{3})}^{2} - 13 x^{3} + 40 = 0$

Step 1.2

Let $u = x^{3}$ . Substitute $u$ for all occurrences of $x^{3}$ .

$u^{2} - 13 u + 40 = 0$

Step 1.3

Factor $u^{2} - 13 u + 40$ using the AC method.

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Step 1.3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $40$ and whose sum is $- 13$ .

$- 8, - 5$

Step 1.3.2

Write the factored form using these integers.

$(u - 8) (u - 5) = 0$

Step 1.4

Replace all occurrences of $u$ with $x^{3}$ .

$(x^{3} - 8) (x^{3} - 5) = 0$

Step 2

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

$x^{3} - 8 = 0$

$x^{3} - 5 = 0$

Step 3

Set $x^{3} - 8$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $x^{3} - 8$ equal to $0$ .

$x^{3} - 8 = 0$

Step 3.2

Solve $x^{3} - 8 = 0$ for $x$ .

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Step 3.2.1

Add $8$ to both sides of the equation.

$x^{3} = 8$

Step 3.2.2

Subtract $8$ from both sides of the equation.

$x^{3} - 8 = 0$

Step 3.2.3

Factor the left side of the equation.

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Step 3.2.3.1

Rewrite $8$ as $2^{3}$ .

$x^{3} - 2^{3} = 0$

Step 3.2.3.2

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 $b = 2$ .

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

Step 3.2.3.3

Simplify.

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Step 3.2.3.3.1

Move $2$ to the left of $x$ .

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

Step 3.2.3.3.2

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

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

Step 3.2.4

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

$x - 2 = 0$

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

Step 3.2.5

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

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Step 3.2.5.1

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

$x - 2 = 0$

Step 3.2.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 3.2.6

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

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Step 3.2.6.1

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

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

Step 3.2.6.2

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

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Step 3.2.6.2.1

Use the quadratic formula to find the solutions.

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

Step 3.2.6.2.2

Substitute the values $a = 1$ , $b = 2$ , and $c = 4$ into the quadratic formula and solve for $x$ .

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

Step 3.2.6.2.3

Simplify.

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Step 3.2.6.2.3.1

Simplify the numerator.

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Step 3.2.6.2.3.1.1

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

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

Step 3.2.6.2.3.1.2

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

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Step 3.2.6.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 3.2.6.2.3.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 3.2.6.2.3.1.3

Subtract $16$ from $4$ .

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

Step 3.2.6.2.3.1.4

Rewrite $- 12$ as $- 1 (12)$ .

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

Step 3.2.6.2.3.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

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

Step 3.2.6.2.3.1.6

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

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

Step 3.2.6.2.3.1.7

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

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Step 3.2.6.2.3.1.7.1

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 3.2.6.2.3.1.7.2

Rewrite $4$ as $2^{2}$ .

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

Step 3.2.6.2.3.1.8

Pull terms out from under the radical.

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

Step 3.2.6.2.3.1.9

Move $2$ to the left of $i$ .

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

Step 3.2.6.2.3.2

Multiply $2$ by $1$ .

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

Step 3.2.6.2.3.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$x = - 1 \pm i \sqrt{3}$

Step 3.2.6.2.4

The final answer is the combination of both solutions.

$x = - 1 + i \sqrt{3}, - 1 - i \sqrt{3}$

Step 3.2.7

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

$x = 2, - 1 + i \sqrt{3}, - 1 - i \sqrt{3}$

Step 4

Set $x^{3} - 5$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x^{3} - 5$ equal to $0$ .

$x^{3} - 5 = 0$

Step 4.2

Solve $x^{3} - 5 = 0$ for $x$ .

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Step 4.2.1

Add $5$ to both sides of the equation.

$x^{3} = 5$

Step 4.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{5}$

Step 5

The final solution is all the values that make $(x^{3} - 8) (x^{3} - 5) = 0$ true.

$x = 2, - 1 + i \sqrt{3}, - 1 - i \sqrt{3}, \sqrt[3]{5}$