Find all the roots of the equation $\displaystyle 4x^{4}-24x^{3}+57x^{2}+18x-45= 0$, if one of them is $\displaystyle 3+i\sqrt{6}.$
- $\displaystyle 3\pm i\sqrt{6};\> \pm \dfrac{\sqrt{3}}{2}$
- $\displaystyle \pm3+ i\sqrt{6};\> \pm \dfrac{\sqrt{3}}{2} $
- $\displaystyle 3\pm i\sqrt{6};\> \dfrac{\sqrt{3}}{2}$
- $\displaystyle \pm3+ i\sqrt{6};\> - \dfrac{\sqrt{3}}{2} $
Since coefficients are real, if 3 + i*sqrt(6) is a root, 3 - i*sqrt(6) is also a root. The quadratic factor is (x - (3 + i*sqrt(6)))(x - (3 - i*sqrt(6))) = (x - 3)^2 + 6 = x^2 - 6x + 15. Dividing 4x^4 - 24x^3 + 57x^2 + 18x - 45 by x^2 - 6x + 15 gives 4x^2 - 3 = 0. Roots are x^2 = 3/4, x = +/- sqrt(3)/2.
Because the coefficients of the polynomial are real, complex roots must occur in conjugate pairs, so since 3 + i*sqrt(6) is a root, 3 - i*sqrt(6) must also be a root. We can factor out the quadratic component corresponding to these roots, which is (x - (3 + i*sqrt(6)))(x - (3 - i*sqrt(6))), simplifying to (x - 3)^2 + 6, or x^2 - 6x + 15. Dividing the original quartic polynomial 4x^4 - 24x^3 + 57x^2 + 18x - 45 by this quadratic factor yields the other quadratic factor, 4x^2 - 3. Setting 4x^2 - 3 equal to 0 gives the remaining roots as plus or minus sqrt(3)/2. The result is 3 plus or minus i*sqrt(6) and plus or minus sqrt(3)/2.