Cubic Equation Fed62D

1. **State the problem:** We need to solve the cubic equation $$x^{3}-3x^{2}-9x+11=0$$ for $x$. 2. **Recall the general approach:** For cubic equations, we can try to find rational roots using the Rational Root Theorem or use substitution methods. 3. **Try possible rational roots:** Possible roots are factors of the constant term 11, i.e., $\pm1, \pm11$. 4. **Test $x=1$:** $$1^{3}-3(1)^{2}-9(1)+11=1-3-9+11=0$$ So, $x=1$ is a root. 5. **Divide the cubic by $(x-1)$:** Using polynomial division or synthetic division: $$\frac{x^{3}-3x^{2}-9x+11}{x-1} = x^{2}-2x-11$$ 6. **Solve the quadratic:** $$x^{2}-2x-11=0$$ Use the quadratic formula: $$x=\frac{2 \pm \sqrt{(-2)^{2}-4(1)(-11)}}{2(1)}=\frac{2 \pm \sqrt{4+44}}{2}=\frac{2 \pm \sqrt{48}}{2}$$ 7. **Simplify the square root:** $$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$$ 8. **Final roots from quadratic:** $$x=\frac{2 \pm 4\sqrt{3}}{2} = 1 \pm 2\sqrt{3}$$ 9. **All solutions:** $$x=1, \quad x=1+2\sqrt{3}, \quad x=1-2\sqrt{3}$$