Trigonometry Examples

$x^{4} + 6 x^{3} + 2 x^{2} + 54 x - 63$

Step 1

Set $x^{4} + 6 x^{3} + 2 x^{2} + 54 x - 63$ equal to $0$ .

$x^{4} + 6 x^{3} + 2 x^{2} + 54 x - 63 = 0$

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.

$6 x^{3} + 54 x + x^{4} + 2 x^{2} - 63 = 0$

Step 2.1.2

Factor $6 x$ out of $6 x^{3} + 54 x$ .

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Step 2.1.2.1

Factor $6 x$ out of $6 x^{3}$ .

$6 x (x^{2}) + 54 x + x^{4} + 2 x^{2} - 63 = 0$

Step 2.1.2.2

Factor $6 x$ out of $54 x$ .

$6 x (x^{2}) + 6 x (9) + x^{4} + 2 x^{2} - 63 = 0$

Step 2.1.2.3

Factor $6 x$ out of $6 x (x^{2}) + 6 x (9)$ .

$6 x (x^{2} + 9) + x^{4} + 2 x^{2} - 63 = 0$

Step 2.1.3

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

$6 x (x^{2} + 9) + {(x^{2})}^{2} + 2 x^{2} - 63 = 0$

Step 2.1.4

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

$6 x (x^{2} + 9) + u^{2} + 2 u - 63 = 0$

Step 2.1.5

Factor $u^{2} + 2 u - 63$ using the AC method.

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Step 2.1.5.1

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 $- 63$ and whose sum is $2$ .

$- 7, 9$

Step 2.1.5.2

Write the factored form using these integers.

$6 x (x^{2} + 9) + (u - 7) (u + 9) = 0$

Step 2.1.6

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

$6 x (x^{2} + 9) + (x^{2} - 7) (x^{2} + 9) = 0$

Step 2.1.7

Factor $x^{2} + 9$ out of $6 x (x^{2} + 9) + (x^{2} - 7) (x^{2} + 9)$ .

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Step 2.1.7.1

Factor $x^{2} + 9$ out of $6 x (x^{2} + 9)$ .

$(x^{2} + 9) (6 x) + (x^{2} - 7) (x^{2} + 9) = 0$

Step 2.1.7.2

Factor $x^{2} + 9$ out of $(x^{2} - 7) (x^{2} + 9)$ .

$(x^{2} + 9) (6 x) + (x^{2} + 9) (x^{2} - 7) = 0$

Step 2.1.7.3

Factor $x^{2} + 9$ out of $(x^{2} + 9) (6 x) + (x^{2} + 9) (x^{2} - 7)$ .

$(x^{2} + 9) (6 x + x^{2} - 7) = 0$

Step 2.1.8

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$(x^{2} + 9) (6 u + u^{2} - 7) = 0$

Step 2.1.9

Factor $6 u + u^{2} - 7$ using the AC method.

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Step 2.1.9.1

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 $- 7$ and whose sum is $6$ .

$- 1, 7$

Step 2.1.9.2

Write the factored form using these integers.

$(x^{2} + 9) ((u - 1) (u + 7)) = 0$

Step 2.1.10

Factor.

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Step 2.1.10.1

Replace all occurrences of $u$ with $x$ .

$(x^{2} + 9) ((x - 1) (x + 7)) = 0$

Step 2.1.10.2

Remove unnecessary parentheses.

$(x^{2} + 9) (x - 1) (x + 7) = 0$

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 $0$ .

$x^{2} + 9 = 0$

$x - 1 = 0$

$x + 7 = 0$

Step 2.3

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

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Step 2.3.1

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

$x^{2} + 9 = 0$

Step 2.3.2

Solve $x^{2} + 9 = 0$ for $x$ .

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Step 2.3.2.1

Subtract $9$ from both sides of the equation.

$x^{2} = - 9$

Step 2.3.2.2

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

$x = \pm \sqrt{- 9}$

Step 2.3.2.3

Simplify $\pm \sqrt{- 9}$ .

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Step 2.3.2.3.1

Rewrite $- 9$ as $- 1 (9)$ .

$x = \pm \sqrt{- 1 (9)}$

Step 2.3.2.3.2

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{9}$

Step 2.3.2.3.3

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

$x = \pm i \cdot \sqrt{9}$

Step 2.3.2.3.4

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

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

Step 2.3.2.3.5

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 3$

Step 2.3.2.3.6

Move $3$ to the left of $i$ .

$x = \pm 3 i$

Step 2.3.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.3.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 3 i$

Step 2.3.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3 i$

Step 2.3.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3 i, - 3 i$

Step 2.4

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 2.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.5

Set $x + 7$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Step 2.5.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 2.6

The final solution is all the values that make $(x^{2} + 9) (x - 1) (x + 7) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 3 i$ (Multiplicity of $1$ )

$x = - 3 i$ (Multiplicity of $1$ )

$x = 1$ (Multiplicity of $1$ )

$x = - 7$ (Multiplicity of $1$ )

$x = 3 i$ (Multiplicity of $1$ )

$x = - 3 i$ (Multiplicity of $1$ )

$x = 1$ (Multiplicity of $1$ )

$x = - 7$ (Multiplicity of $1$ )

Step 3