Precalculus Examples

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

Step 1

Factor the left side of the equation.

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Step 1.1

Factor out the greatest common factor from each group.

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Step 1.1.1

Group the first two terms and the last two terms.

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

Step 1.1.2

Factor out the greatest common factor (GCF) from each group.

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

Step 1.2

Factor the polynomial by factoring out the greatest common factor, $x - 3$ .

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

Step 2

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

$x - 3 = 0$

$x^{2} + 2 = 0$

Step 3

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

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Step 3.1

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

$x - 3 = 0$

Step 3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 4

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

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Step 4.1

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

$x^{2} + 2 = 0$

Step 4.2

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

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Step 4.2.1

Subtract $2$ from both sides of the equation.

$x^{2} = - 2$

Step 4.2.2

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

$x = \pm \sqrt{- 2}$

Step 4.2.3

Simplify $\pm \sqrt{- 2}$ .

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Step 4.2.3.1

Rewrite $- 2$ as $- 1 (2)$ .

$x = \pm \sqrt{- 1 (2)}$

Step 4.2.3.2

Rewrite $\sqrt{- 1 (2)}$ as $\sqrt{- 1} \cdot \sqrt{2}$ .

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

Step 4.2.3.3

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

$x = \pm i \sqrt{2}$

Step 4.2.4

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

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Step 4.2.4.1

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

$x = i \sqrt{2}$

Step 4.2.4.2

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

$x = - i \sqrt{2}$

Step 4.2.4.3

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

$x = i \sqrt{2}, - i \sqrt{2}$

Step 5

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

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