Precalculus Examples

$x (x - 1) = 30$

Step 1

Subtract $30$ from both sides of the equation.

$x (x - 1) - 30 = 0$

Step 2

Simplify each term.

Tap for more steps...

Step 2.1

Apply the distributive property.

$x \cdot x + x \cdot - 1 - 30 = 0$

Step 2.2

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 - 30 = 0$

Step 2.3

Move $- 1$ to the left of $x$ .

$x^{2} - 1 \cdot x - 30 = 0$

Step 2.4

Rewrite $- 1 x$ as $- x$ .

$x^{2} - x - 30 = 0$

Step 3

Factor $x^{2} - x - 30$ using the AC method.

Tap for more steps...

Step 3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 30$ and whose sum is $- 1$ .

$- 6, 5$

Step 3.2

Write the factored form using these integers.

$(x - 6) (x + 5) = 0$

Step 4

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

$x - 6 = 0$

$x + 5 = 0$

Step 5

Set $x - 6$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 5.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 5.2

Add $6$ to both sides of the equation.

$x = 6$

Step 6

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

Tap for more steps...

Step 6.1

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

$x + 5 = 0$

Step 6.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 7

The final solution is all the values that make $(x - 6) (x + 5) = 0$ true.

$x = 6, - 5$