Signs Under Pressure

To factor $x^{2}-x-12$, find two numbers whose product is -12 and whose sum is -1. They are -4 and 3, so $x^{2}-x-12 = \left(x-4\right)\left(x+3\right)$. The equation becomes $\left(x-4\right)\left(x+3\right) = 0$.

Set each factor equal to zero: $x = 4$ or $x = -3$. Notice that the signs change twice. The pair inside the parentheses is -4 and 3. The solutions have the opposite signs, 4 and -3.

First check the signs in the factored form. Then solve each factor equation separately. $x-4 = 0$ gives $x = 4$, and $x+3 = 0$ gives $x = -3$.