1. **State the problem:** Solve the cubic equation $$x^3 - 5x^2 - 12x + 60 = 0$$ given that $$x = \pm 2\sqrt{3}$$ are solutions. 2. **Use the factored form:** Since $$x = 2\sqrt{3}$$ and $$x = -2\sqrt{3}$$ are roots, the polynomial must be divisible by $$\left(x - 2\sqrt{3}\right)\left(x + 2\sqrt{3}\right) = x^2 - (2\sqrt{3})^2 = x^2 - 12$$. 3. **Divide the cubic by the quadratic factor:** We perform polynomial division of $$x^3 - 5x^2 - 12x + 60$$ by $$x^2 - 12$$. Set up division: $$\frac{x^3 - 5x^2 - 12x + 60}{x^2 - 12}$$ - Divide leading terms: $$\frac{x^3}{x^2} = x$$ - Multiply divisor by $$x$$: $$x(x^2 - 12) = x^3 - 12x$$ - Subtract: $$\left(x^3 - 5x^2 - 12x + 60\right) - \left(x^3 - 12x\right) = -5x^2 + 0x + 60$$ - Divide leading terms: $$\frac{-5x^2}{x^2} = -5$$ - Multiply divisor by $$-5$$: $$-5(x^2 - 12) = -5x^2 + 60$$ - Subtract: $$\left(-5x^2 + 0x + 60\right) - \left(-5x^2 + 60\right) = 0$$ So the quotient is $$x - 5$$ and remainder is 0. 4. **Write the factorization:** $$x^3 - 5x^2 - 12x + 60 = (x^2 - 12)(x - 5)$$ 5. **Find all solutions:** - From $$x^2 - 12 = 0$$, we get $$x = \pm \sqrt{12} = \pm 2\sqrt{3}$$ (given). - From $$x - 5 = 0$$, we get $$x = 5$$. 6. **Final answer:** $$\boxed{x = -2\sqrt{3}, \quad x = 2\sqrt{3}, \quad x = 5}$$