Cubic Equation 9Eb872

1. **State the problem:** Solve the cubic equation $$x^{3}-3x^{2}-9x+11=0$$. 2. **Recall the general approach:** For cubic equations, one method is to try to find rational roots using the Rational Root Theorem or attempt substitution to simplify. 3. **Try possible rational roots:** Factors of the constant term 11 are $\pm1, \pm11$. Test these values: - For $x=1$: $$1^{3}-3(1)^{2}-9(1)+11=1-3-9+11=0$$, so $x=1$ is a root. 4. **Divide the cubic by $(x-1)$:** Use polynomial division or synthetic division: $$\frac{x^{3}-3x^{2}-9x+11}{x-1} = x^{2}-2x-11$$ 5. **Solve the quadratic equation:** $$x^{2}-2x-11=0$$ Use the quadratic formula: $$x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^{2}-4(1)(-11)}}{2} = \frac{2 \pm \sqrt{4+44}}{2} = \frac{2 \pm \sqrt{48}}{2}$$ Simplify $$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$$: $$x=\frac{2 \pm 4\sqrt{3}}{2} = 1 \pm 2\sqrt{3}$$ 6. **Final solutions:** $$x=1, \quad x=1+2\sqrt{3}, \quad x=1-2\sqrt{3}$$