Q.Find the exact number of real zeros and imaginary of the polynomial $x^9+9x^7+7x^5+5x^3+3x$.

Step 1. Factor out the common $x$. $x^9+9x^7+7x^5+5x^3+3x=x\big(x^8+9x^6+7x^4+5x^2+3\big)$.

Step 2. Analyse the degree-8 factor $Q(x)=x^8+9x^6+7x^4+5x^2+3$. Every term has a positive coefficient and an even power, so for every real $x$ (including $x=0$), $Q(x)=x^8+9x^6+7x^4+5x^2+3\ge3>0$ — $Q(x)$ is never zero for real $x$.

Step 3. Confirm via Descartes too. $Q(x)$'s coefficients are all $+$: $0$ sign changes, so $0$ positive roots. $Q(-x)=Q(x)$ (all even powers): also $0$ sign changes, so $0$ negative roots. Consistent with Step 2 — $Q$ has no real roots at all, so all $8$ of its roots are non-real (imaginary). …

Factor out the common power of $x$, then note an all-positive-coefficient, even-powers-only p …

• Trying to apply Descartes to the ORIGINAL degree-9 polynomial without first factoring out $x$ (the odd-degree $x^9$ term makes the direct sign count less immediately conclusive) …