Precalculus Examples

$f (x) = x^{4} - 2 x^{3} - 15 x^{2} + 18 x + 54$

Step 1

Set $x^{4} - 2 x^{3} - 15 x^{2} + 18 x + 54$ equal to $0$ .

$x^{4} - 2 x^{3} - 15 x^{2} + 18 x + 54 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Regroup terms.

$- 2 x^{3} + 18 x + x^{4} - 15 x^{2} + 54 = 0$

Step 2.1.2

Factor $- 2 x$ out of $- 2 x^{3} + 18 x$ .

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Step 2.1.2.1

Factor $- 2 x$ out of $- 2 x^{3}$ .

$- 2 x \cdot x^{2} + 18 x + x^{4} - 15 x^{2} + 54 = 0$

Step 2.1.2.2

Factor $- 2 x$ out of $18 x$ .

$- 2 x \cdot x^{2} - 2 x \cdot - 9 + x^{4} - 15 x^{2} + 54 = 0$

Step 2.1.2.3

Factor $- 2 x$ out of $- 2 x (x^{2}) - 2 x (- 9)$ .

$- 2 x (x^{2} - 9) + x^{4} - 15 x^{2} + 54 = 0$

Step 2.1.3

Rewrite $9$ as $3^{2}$ .

$- 2 x (x^{2} - 3^{2}) + x^{4} - 15 x^{2} + 54 = 0$

Step 2.1.4

Factor.

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Step 2.1.4.1

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

$- 2 x ((x + 3) (x - 3)) + x^{4} - 15 x^{2} + 54 = 0$

Step 2.1.4.2

Remove unnecessary parentheses.

$- 2 x (x + 3) (x - 3) + x^{4} - 15 x^{2} + 54 = 0$

Step 2.1.5

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$- 2 x (x + 3) (x - 3) + {(x^{2})}^{2} - 15 x^{2} + 54 = 0$

Step 2.1.6

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- 2 x (x + 3) (x - 3) + u^{2} - 15 u + 54 = 0$

Step 2.1.7

Factor $u^{2} - 15 u + 54$ using the AC method.

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Step 2.1.7.1

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 $54$ and whose sum is $- 15$ .

$- 9, - 6$

Step 2.1.7.2

Write the factored form using these integers.

$- 2 x (x + 3) (x - 3) + (u - 9) (u - 6) = 0$

Step 2.1.8

Replace all occurrences of $u$ with $x^{2}$ .

$- 2 x (x + 3) (x - 3) + (x^{2} - 9) (x^{2} - 6) = 0$

Step 2.1.9

Rewrite $9$ as $3^{2}$ .

$- 2 x (x + 3) (x - 3) + (x^{2} - 3^{2}) (x^{2} - 6) = 0$

Step 2.1.10

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

$- 2 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} - 6) = 0$

Step 2.1.11

Factor $(x + 3) (x - 3)$ out of $- 2 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} - 6)$ .

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Step 2.1.11.1

Factor $(x + 3) (x - 3)$ out of $- 2 x (x + 3) (x - 3)$ .

$(x + 3) (x - 3) (- 2 x) + (x + 3) (x - 3) (x^{2} - 6) = 0$

Step 2.1.11.2

Factor $(x + 3) (x - 3)$ out of $(x + 3) (x - 3) (- 2 x) + (x + 3) (x - 3) (x^{2} - 6)$ .

$(x + 3) (x - 3) (- 2 x + x^{2} - 6) = 0$

Step 2.1.12

Reorder terms.

$(x + 3) (x - 3) (x^{2} - 2 x - 6) = 0$

Step 2.2

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

$x + 3 = 0$

$x - 3 = 0$

$x^{2} - 2 x - 6 = 0$

Step 2.3

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.3.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.4

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

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Step 2.4.1

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

$x - 3 = 0$

Step 2.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.5

Set $x^{2} - 2 x - 6$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x^{2} - 2 x - 6$ equal to $0$ .

$x^{2} - 2 x - 6 = 0$

Step 2.5.2

Solve $x^{2} - 2 x - 6 = 0$ for $x$ .

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Step 2.5.2.1

Use the quadratic formula to find the solutions.

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

Step 2.5.2.2

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 6)}}{2 \cdot 1}$

Step 2.5.2.3

Simplify.

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Step 2.5.2.3.1

Simplify the numerator.

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Step 2.5.2.3.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 6}}{2 \cdot 1}$

Step 2.5.2.3.1.2

Multiply $- 4 \cdot 1 \cdot - 6$ .

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Step 2.5.2.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 6}}{2 \cdot 1}$

Step 2.5.2.3.1.2.2

Multiply $- 4$ by $- 6$ .

$x = \frac{2 \pm \sqrt{4 + 24}}{2 \cdot 1}$

Step 2.5.2.3.1.3

Add $4$ and $24$ .

$x = \frac{2 \pm \sqrt{28}}{2 \cdot 1}$

Step 2.5.2.3.1.4

Rewrite $28$ as $2^{2} \cdot 7$ .

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Step 2.5.2.3.1.4.1

Factor $4$ out of $28$ .

$x = \frac{2 \pm \sqrt{4 (7)}}{2 \cdot 1}$

Step 2.5.2.3.1.4.2

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 1}$

Step 2.5.2.3.1.5

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{7}}{2 \cdot 1}$

Step 2.5.2.3.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{7}}{2}$

Step 2.5.2.3.3

Simplify $\frac{2 \pm 2 \sqrt{7}}{2}$ .

$x = 1 \pm \sqrt{7}$

Step 2.5.2.4

The final answer is the combination of both solutions.

$x = 1 + \sqrt{7}, 1 - \sqrt{7}$

Step 2.6

The final solution is all the values that make $(x + 3) (x - 3) (x^{2} - 2 x - 6) = 0$ true.

$x = - 3, 3, 1 + \sqrt{7}, 1 - \sqrt{7}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = - 3, 3, 1 + \sqrt{7}, 1 - \sqrt{7}$

Decimal Form:

$x = - 3, 3, 3.64575131 \dots, - 1.64575131 \dots$

Step 4