Q.Discuss the maximum possible number of positive and negative zeros of the polynomials $x^2-5x+6$ and $x^2-5x+16$. Also draw rough sketch of the graphs.

So $p\le s$, and $p$ has the same parity as $s$ (if $s=4$, then $p\in\{0,2,4\}$, never $1$ or $3$). Applying the same rule to $P(-x)$ bounds the number of negative roots (since a negative root of $P$ is a positive root of $P(-x)$).

Combining both bounds — a lower bound on non-real roots. For a degree-$n$ polynomial, with $m$ = sign changes of $P(x)$ and $k$ = sign changes of $P(-x)$: since the equation has exactly $n$ roots in all (Fundamental Theorem of Algebra) and at most $m+k$ of them can be real,

When is the bound tight? Sometimes it is attained exactly (e.g. $(x-1)^n=0$ has all $n$ roots positive with multiplicity $n$, matching its $n$ sign changes exactly, and $0$ sign changes in $P(-x)$ correctly rules out any negative root); other times it is a strict inequality, and only the parity fact narrows things further — e.g. combined with "coefficients sum to zero forces $1$ to be a root" (see Polynomials with Additional Conditions), a bound of "at most $2$ positive roots" can be sharpened to "exactly $2$" once one positive root is confirmed directly.

Worked illustration. $P(x)=9x^9+2x^5-x^4-7x^2+2$: signs $+,+,-,-,+$ give $2$ sign changes (at most $2$ positive roots); $P(-x)=-9x^9-2x^5-x^4-7x^2+2$ has signs $-,-,-,-,+$, giving $1$ sign change (at most $1$ negative root). Since $0$ is not a root (constant term $2\ne0$), at most $2+1=3$ of the $9$ roots are real — so at least $6$ are non-real, proved with no solving at all.