1. **State the problem:** Solve the equation by factoring: $$4x(3x + 1) = 1$$ 2. **Rewrite the equation:** Expand the left side: $$4x \cdot 3x + 4x \cdot 1 = 1$$ which simplifies to $$12x^2 + 4x = 1$$ 3. **Bring all terms to one side:** $$12x^2 + 4x - 1 = 0$$ 4. **Factor the quadratic equation:** We look for two numbers that multiply to $$12 \times (-1) = -12$$ and add to $$4$$. These numbers are $$6$$ and $$-2$$. Rewrite the middle term: $$12x^2 + 6x - 2x - 1 = 0$$ Group terms: $$(12x^2 + 6x) - (2x + 1) = 0$$ Factor each group: $$6x(2x + 1) - 1(2x + 1) = 0$$ Factor out the common binomial: $$(6x - 1)(2x + 1) = 0$$ 5. **Solve each factor for zero:** $$6x - 1 = 0 \implies 6x = 1 \implies x = \frac{1}{6}$$ $$2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}$$ 6. **Final answer:** $$x = \frac{1}{6}, -\frac{1}{2}$$