Pre-Algebra Examples

$16 = (x + 3) (x - 3)$

Step 1

Rewrite the equation as $(x + 3) (x - 3) = 16$ .

$(x + 3) (x - 3) = 16$

Step 2

Simplify $(x + 3) (x - 3)$ .

Tap for more steps...

Step 2.1

Expand $(x + 3) (x - 3)$ using the FOIL Method.

Tap for more steps...

Step 2.1.1

Apply the distributive property.

$x (x - 3) + 3 (x - 3) = 16$

Step 2.1.2

Apply the distributive property.

$x \cdot x + x \cdot - 3 + 3 (x - 3) = 16$

Step 2.1.3

Apply the distributive property.

$x \cdot x + x \cdot - 3 + 3 x + 3 \cdot - 3 = 16$

Step 2.2

Simplify terms.

Tap for more steps...

Step 2.2.1

Combine the opposite terms in $x \cdot x + x \cdot - 3 + 3 x + 3 \cdot - 3$ .

Tap for more steps...

Step 2.2.1.1

Reorder the factors in the terms $x \cdot - 3$ and $3 x$ .

$x \cdot x - 3 x + 3 x + 3 \cdot - 3 = 16$

Step 2.2.1.2

Add $- 3 x$ and $3 x$ .

$x \cdot x + 0 + 3 \cdot - 3 = 16$

Step 2.2.1.3

Add $x \cdot x$ and $0$ .

$x \cdot x + 3 \cdot - 3 = 16$

Step 2.2.2

Simplify each term.

Tap for more steps...

Step 2.2.2.1

Multiply $x$ by $x$ .

$x^{2} + 3 \cdot - 3 = 16$

Step 2.2.2.2

Multiply $3$ by $- 3$ .

$x^{2} - 9 = 16$

Step 3

Move all terms not containing $x$ to the right side of the equation.

Tap for more steps...

Step 3.1

Add $9$ to both sides of the equation.

$x^{2} = 16 + 9$

Step 3.2

Add $16$ and $9$ .

$x^{2} = 25$

Step 4

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

$x = \pm \sqrt{25}$

Step 5

Simplify $\pm \sqrt{25}$ .

Tap for more steps...

Step 5.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Step 5.2

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

$x = \pm 5$

Step 6

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

Tap for more steps...

Step 6.1

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

$x = 5$

Step 6.2

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

$x = - 5$

Step 6.3

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

$x = 5, - 5$