Solve Equations 1Db47C

1. **State the problem:** We need to solve the equation $$(x - 10)(x + 20) = 0$$. 2. **Recall the zero product property:** If the product of two factors is zero, then at least one of the factors must be zero. That is, $$ (x - 10)(x + 20) = 0 \implies x - 10 = 0 \text{ or } x + 20 = 0 $$ 3. **Solve each equation separately:** - For $$x - 10 = 0$$, add 10 to both sides: $$ x = 10 $$ - For $$x + 20 = 0$$, subtract 20 from both sides: $$ x = -20 $$ 4. **Final solution:** The solutions to the equation are $$x = 10$$ and $$x = -20$$. --- Now, solve the equation $$(3x + 4)(x - 4) = 0$$. 5. **Apply zero product property:** $$ (3x + 4)(x - 4) = 0 \implies 3x + 4 = 0 \text{ or } x - 4 = 0 $$ 6. **Solve each equation:** - For $$3x + 4 = 0$$, subtract 4 from both sides: $$ 3x = -4 $$ Then divide both sides by 3: $$ x = \frac{\cancel{3}x}{\cancel{3}} = \frac{-4}{3} $$ - For $$x - 4 = 0$$, add 4 to both sides: $$ x = 4 $$ 7. **Final solution:** The solutions to the equation are $$x = -\frac{4}{3}$$ and $$x = 4$$.