Algebra Examples

$f (x) = (x + 5) (x^{2} - 4)$

Step 1

Set $(x + 5) (x^{2} - 4)$ equal to $0$ .

$(x + 5) (x^{2} - 4) = 0$

Step 2

Solve for $x$ .

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Step 2.1

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

$x + 5 = 0$

$x^{2} - 4 = 0$

Step 2.2

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 2.2.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 2.2.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.3

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

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Step 2.3.1

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

$x^{2} - 4 = 0$

Step 2.3.2

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

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Step 2.3.2.1

Add $4$ to both sides of the equation.

$x^{2} = 4$

Step 2.3.2.2

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

$x = \pm \sqrt{4}$

Step 2.3.2.3

Simplify $\pm \sqrt{4}$ .

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Step 2.3.2.3.1

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

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

Step 2.3.2.3.2

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

$x = \pm 2$

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.

$x = 2$

Step 2.3.2.4.2

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

$x = - 2$

Step 2.3.2.4.3

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

$x = 2, - 2$

Step 2.4

The final solution is all the values that make $(x + 5) (x^{2} - 4) = 0$ true.

$x = - 5, 2, - 2$

Step 3