Calculus Examples

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

Step 1

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

$x - 3 = 0$

$x^{2} + 4 i x - 4 i x - 16 i^{2} = 0$

Step 2

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

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Step 2.1

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

$x - 3 = 0$

Step 2.2

Add $3$ to both sides of the equation.

$x = 3$

Step 3

Set $x^{2} + 4 i x - 4 i x - 16 i^{2}$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $x^{2} + 4 i x - 4 i x - 16 i^{2}$ equal to $0$ .

$x^{2} + 4 i x - 4 i x - 16 i^{2} = 0$

Step 3.2

Solve $x^{2} + 4 i x - 4 i x - 16 i^{2} = 0$ for $x$ .

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Step 3.2.1

Simplify $x^{2} + 4 i x - 4 i x - 16 i^{2}$ .

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Step 3.2.1.1

Combine the opposite terms in $x^{2} + 4 i x - 4 i x - 16 i^{2}$ .

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Step 3.2.1.1.1

Subtract $4 i x$ from $4 i x$ .

$x^{2} + 0 - 16 i^{2} = 0$

Step 3.2.1.1.2

Add $x^{2}$ and $0$ .

$x^{2} - 16 i^{2} = 0$

Step 3.2.1.2

Simplify each term.

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Step 3.2.1.2.1

Rewrite $i^{2}$ as $- 1$ .

$x^{2} - 16 \cdot - 1 = 0$

Step 3.2.1.2.2

Multiply $- 16$ by $- 1$ .

$x^{2} + 16 = 0$

Step 3.2.2

Subtract $16$ from both sides of the equation.

$x^{2} = - 16$

Step 3.2.3

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

$x = \pm \sqrt{- 16}$

Step 3.2.4

Simplify $\pm \sqrt{- 16}$ .

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Step 3.2.4.1

Rewrite $- 16$ as $- 1 (16)$ .

$x = \pm \sqrt{- 1 (16)}$

Step 3.2.4.2

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

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

Step 3.2.4.3

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

$x = \pm i \cdot \sqrt{16}$

Step 3.2.4.4

Rewrite $16$ as $4^{2}$ .

$x = \pm i \cdot \sqrt{4^{2}}$

Step 3.2.4.5

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

$x = \pm i \cdot 4$

Step 3.2.4.6

Move $4$ to the left of $i$ .

$x = \pm 4 i$

Step 3.2.5

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

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Step 3.2.5.1

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

$x = 4 i$

Step 3.2.5.2

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

$x = - 4 i$

Step 3.2.5.3

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

$x = 4 i, - 4 i$

Step 4

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

$x = 3, 4 i, - 4 i$