Question

If one root of the equation $x^4+4x^3+6x^2+4x+5$ is $\sqrt{-1}$ . Find the sum of the squares of other three roots.

Answer 1

One root = $\sqrt{-1}$ = i

Since, complex root occurs in conjugate one more root should be - i

x = i

$x^2=i^2=-1$

($x^2+1$) = 0

Divide the given $x^4+4x^3+6x^2+4x+5=0by{x}^2+1$

$\begin{array}{cc}& x^2+4x+5\\ x^2+1& x^4+4x^3+6x^2+4x+5\\ & x^4+x^2\\ & --\\ & 4x^3+5x^2\\ & 4x^3+4x\\ & --\\ & 5x^2+5\\ & 5x^2+5\\ & 0\end{array}$