Question

Solve $x^4-4x^2+8x+35=0$, given that $2+i\sqrt{3}$ is a root.

Answer 1

Since $2+i\sqrt{3}$ is a root and the coefficients are real, the second root has to be $2-i\sqrt{3}$

Thus, $(x-2-i\sqrt{3})(x-2+i\sqrt{3})=x^2-2x+ix\sqrt{3}-2x+4-i2\sqrt{3}-ix\sqrt{3}+i2\sqrt{3}+3$

$=x^2-4x+7$ is a root of the given equation.

Dividing the original equation by $x^2-4x+7$, we have

$(x^4-4x^2+8x+35)/(x^2-4x+7)=x^2+4x+5$

Thus, we obtain the other root as $x^2+4x+5$

Again factorizing this root, we get $x=\frac{-4\pm \sqrt{16-20}}{2}$

$=\frac{-4\pm 2i}{2}=-2\pm i$

The four roots are therefore $2+i\sqrt{3},2-i\sqrt{3},-2+i,-2-i$