Question

lf $2-i\sqrt{3}$ is a root of $x^4-4x^2+8x+35=0$ then the other roots are

• A. $2+i\sqrt{3},2+i,-2+i$
• B. $2+i\sqrt{3},-2+i,2-i$
• C. $2+i\sqrt{3},-2+i,-2-i$
• D. $2+i\sqrt{3},2+i,-2-i$

Answer 1

The correct option is C $2+i\sqrt{3},-2+i,-2-i$

Since all the co-efficients are real, we have $2-i\sqrt{3}$ as well as its conjugate $2+i\sqrt{3}$ as the roots.
Thus dividing the polynomial by $(x-2-i\sqrt{3})(x-2+i\sqrt{3})$ i.e. $x^2-4x+7,$ we get the quotient as $x^2+4x+5.$
Thus solving this quadratic equation, we get the other 2 roots as $-2+i$ and $-2-i.$