Precalculus Examples

$5 x^{3} - x^{2} + 5 x - 1$

Step 1

Set $5 x^{3} - x^{2} + 5 x - 1$ equal to $0$ .

$5 x^{3} - x^{2} + 5 x - 1 = 0$

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

Factor out the greatest common factor from each group.

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Step 2.1.1.1

Group the first two terms and the last two terms.

$(5 x^{3} - x^{2}) + 5 x - 1 = 0$

Step 2.1.1.2

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

$x^{2} (5 x - 1) + 1 (5 x - 1) = 0$

Step 2.1.2

Factor the polynomial by factoring out the greatest common factor, $5 x - 1$ .

$(5 x - 1) (x^{2} + 1) = 0$

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 $0$ .

$5 x - 1 = 0$

$x^{2} + 1 = 0$

Step 2.3

Set $5 x - 1$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $5 x - 1$ equal to $0$ .

$5 x - 1 = 0$

Step 2.3.2

Solve $5 x - 1 = 0$ for $x$ .

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Step 2.3.2.1

Add $1$ to both sides of the equation.

$5 x = 1$

Step 2.3.2.2

Divide each term in $5 x = 1$ by $5$ and simplify.

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Step 2.3.2.2.1

Divide each term in $5 x = 1$ by $5$ .

$\frac{5 x}{5} = \frac{1}{5}$

Step 2.3.2.2.2

Simplify the left side.

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Step 2.3.2.2.2.1

Cancel the common factor of $5$ .

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Step 2.3.2.2.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{1}{5}$

Step 2.3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{5}$

Step 2.4

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Step 2.4.2

Solve $x^{2} + 1 = 0$ for $x$ .

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Step 2.4.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 2.4.2.2

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

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

Step 2.4.2.3

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

$x = \pm i$

Step 2.4.2.4

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

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Step 2.4.2.4.1

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

$x = i$

Step 2.4.2.4.2

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

$x = - i$

Step 2.4.2.4.3

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

$x = i, - i$

Step 2.5

The final solution is all the values that make $(5 x - 1) (x^{2} + 1) = 0$ true.

$x = \frac{1}{5}, i, - i$

Step 3