Q.Show that the equation $x^9-5x^5+4x^4+2x^2+1=0$ has atleast $6$ imaginary solutions.

Step 1. Write $P(x)=x^9-5x^5+4x^4+2x^2+1$. Signs (skipping zero $x^8,x^7,x^6,x^3,x^1$ terms): $+,-,+,+,+$. Changes: $x^9{\to}x^5$ ($+\to-$), $x^5{\to}x^4$ ($-\to+$): 2 changes, so at most $2$ positive roots.

Step 2. Write $P(-x)=-x^9+5x^5+4x^4+2x^2+1$. Signs: $-,+,+,+,+$. Changes: $x^9{\to}x^5$ ($-\to+$): 1 change, so at most $1$ negative root.

Step 3. Bound the real roots. At most $2+1=3$ real roots in total (and clearly $x=0$ is not a root, since the constant term is $1\ne0$). …

Descartes' Rule of Signs — bound the real roots via sign changes in $P(x)$ and $P(-x)$ …

• Miscounting sign changes by not skipping the missing (zero-coefficient) terms correctly …