Question

For the given equation $x^{12}+x^{10}-x^9+x^7-x^6+x^5+x^4-x^2+x-2=0$, how many maximum imaginary roots are possible?___

Answer 1

Let f(x) $=x^{12}+x^{10}-x^9+x^7-x^6+x^5+x^4-x^2+x-2=0$

According to Descartes’ rule, maximum number of positive real roots = number of sign Changes in f(x) = 7

Similarly, Maximum number of negative real roots = number of sign changes in f (-x)

To find f (-x) replace “x” by “-x” in each instance

$f(-x)=x^{12}+x^{10}+x^9-x^7-x^6-x^5+x^4-x^2-x-2=0$

Maximum number of negative real roots = number of sign changes in f (-x) = 3

Zero cannot be a root because the constant part is also involved in the equation.

$\begin{array}{ccc}& & \\ PositiveRealRoots& NegativeRealRoots& ImaginaryRoots\\ 7& 3& 2\\ 5& 3& 4\\ 7& 1& 4\\ 3& 3& 6\\ 5& 1& 6\\ 1& 3& 8\\ 3& 1& 8\\ 1& 1& 10\end{array}$
So, maximum number of imaginary roots = 10.