Precalculus Examples

$x = x^{2} - 12$

Step 1

Subtract $x^{2}$ from both sides of the equation.

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

Step 2

Add $12$ to both sides of the equation.

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

Step 3

Factor the left side of the equation.

Tap for more steps...

Step 3.1

Factor $- 1$ out of $x - x^{2} + 12$ .

Tap for more steps...

Step 3.1.1

Reorder $x$ and $- x^{2}$ .

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

Step 3.1.2

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + x + 12 = 0$

Step 3.1.3

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 12 = 0$

Step 3.1.4

Rewrite $12$ as $- 1 (- 12)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 12 = 0$

Step 3.1.5

Factor $- 1$ out of $- (x^{2}) - 1 (- x)$ .

$- (x^{2} - x) - 1 \cdot - 12 = 0$

Step 3.1.6

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 12)$ .

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

Step 3.2

Factor.

Tap for more steps...

Step 3.2.1

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

Tap for more steps...

Step 3.2.1.1

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 3.2.1.2

Write the factored form using these integers.

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

Step 3.2.2

Remove unnecessary parentheses.

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

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 $x$ .

Tap for more steps...

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 $x$ .

Tap for more steps...

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.

$x = 4, - 3$