Question

Solve the equation. $$ \left(\frac{1}{2} x+2\right)\left(\frac{2}{3} x+6\right)\left(\frac{1}{6} x-1\right)=0 $$

Short answer

The roots of the equation \( \left(\frac{1}{2} x+2\right)\left(\frac{2}{3} x+6\right)\left(\frac{1}{6} x-1\right) = 0 \) are x = -4, x = -9, and x = 6.

Step-by-step solution

Apply the Zero-Product Property

Given that \(\left(\frac{1}{2} x+2\right)\left(\frac{2}{3} x+6\right)\left(\frac{1}{6} x-1\right)=0\), using the Zero-Product Property, set each term to zero independently: \( \frac{1}{2} x+2 = 0, \frac{2}{3} x+6 = 0, \text{ and } \frac{1}{6} x-1 = 0.\)

Solve each equation separately

To find the roots of the original equation, solve the three derived equations independently:1. \( \frac{1}{2} x+2 = 0: Subtract 2 from both sides and then multiply by 2 to isolate x yielding x = -4.2. \( \frac{2}{3} x+6 = 0: Subtract 6 from both sides and then multiply by \(\frac{3}{2}\) to isolate x yielding x = -9.3. \( \frac{1}{6} x-1 = 0: Add 1 to both sides and then multiply by 6 to isolate x yielding x = 6.

Verify the solutions

The solutions found should be plugged back into the original multi-term equation to verify their correctness. Solving the main equation with x = -4, x = -9, and x = 6, you should get zero in all instances because every term results in zero when any of these x-values are used.