Answer
$6$ is an upper bound for the real zeros of $P$
$0$ is a lower bound for the real zeros of $P$
Work Step by Step
Suppose $f$ is a polynomial of degree $n\geq 1$.
If $c>0$ is synthetically divided into $f$ and all of the coefficients of the quotient and remainder are all non-negative values, then $c$ is an upper bound for the real zeros of $f$. That is, there are no real zeros greater than $c$.
If $c<0$ is synthetically divided into $f$ and the coefficients of the quotient and remainder alternate signs, then $c$ is a lower bound for the real zeros of $f$. That is, there are no real zeros less than $c$.
$P(x)=3x^4-17x^3+24x^2-9x+1$; $a=0$, $b=6$
$\begin{array}{lllll}
\underline{6}| & 3&-17 & 24 & -9 & 1& \\
& &18& 6 & 180 & 1026\\
\hline & & & & \\
& 3&1 & 30 & 171 & 1027
\end{array}$
All of the coefficients of the quotient and remainder are all non-negative values. thus, $6$ is an upper bound for the real zeros of $P$.
$\begin{array}{lllll}
\underline{0}| &3& -17 & 24 & -9 & 1\\
& & 0&0 & 0& 0\\
\hline & & & & \\
& 3&-17& 24 & -9 & 1
\end{array}$
The coefficients of the quotient and remainder alternate signs. thus, $0$ is a lower bound for the real zeros of $P$.
