Pre-Algebra Examples

$x (x - 1) = 12$

Step 1

Simplify $x (x - 1)$ .

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Step 1.1

Simplify by multiplying through.

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Step 1.1.1

Apply the distributive property.

$x \cdot x + x \cdot - 1 = 12$

Step 1.1.2

Simplify the expression.

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Step 1.1.2.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 = 12$

Step 1.1.2.2

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

$x^{2} - 1 \cdot x = 12$

Step 1.2

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

$x^{2} - x = 12$

Step 2

Subtract $12$ from both sides of the equation.

$x^{2} - x - 12 = 0$

Step 3

Factor $x^{2} - x - 12$ using the AC method.

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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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Step 3.2

Write the factored form using these integers.

$(x - 4) (x + 3) = 0$

Step 4

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

$x - 4 = 0$

$x + 3 = 0$

Step 5

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 5.2

Add $4$ to both sides of the equation.

$x = 4$

Step 6

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

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Step 6.1

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

$x + 3 = 0$

Step 6.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 7

The final solution is all the values that make $(x - 4) (x + 3) = 0$ true.

$x = 4, - 3$