Solve using the principle of zero products.
-$\left(3 \mathrm{x}-\frac{1}{2}\right)\left(\mathrm{x}+\frac{1}{3}\right)=0$
A) $-\frac{1}{6}, \frac{1}{3}$
B) $\frac{1}{6},-\frac{1}{3}$
C) $-\frac{1}{6},-\frac{1}{3}$
D) $\frac{1}{6}, \frac{1}{3}$

Question

Quizplus · Accepted Answer

The principle of zero products states that if a product of factors equals zero, at least one of the factors must be zero. Solving $3x - \frac{1}{2} = 0$ gives $x = \frac{1}{6}$, and solving $x + \frac{1}{3} = 0$ gives $x = -\frac{1}{3}$.

Solve Using the Principle of Zero Products (3x−12)(x+13)=0\left(3 \mathrm{x}-\frac{1}{2}\right)\left(\mathrm{x}+\frac{1}{3}\right)=0(3x−21​)(x+31​)=0 A) −16,13-\frac{1}{6}, \frac{1}{3}−61​,31​

Solve Using the Principle of Zero Products $\left(3 \mathrm{x}-\frac{1}{2}\right)\left(\mathrm{x}+\frac{1}{3}\right)=0$ A) $-\frac{1}{6}, \frac{1}{3}$