Answer
$5$ is an upper bound for the real zeros of $P$
$-3$ 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 $c$.
$P(x)=x^4-2x^3-9x^2+2x+8$; $a=-3$, $b=5$
$\begin{array}{lllll}
\underline{5}| & 1& -2 & -9 & 2 & 8& \\
& &5& 15 & 30 & 160\\
\hline & & & & \\
& 1& 3 & 6 & 32 & 168
\end{array}$
All of the coefficients of the quotient and remainder are all non-negative values. thus, $5$ is an upper bound for the real zeros of $P$.
$\begin{array}{lllll}
\underline{-3}| &1& -2 & -9 & 2 & 8\\
& & -3 & 15 & -18& 48\\
\hline & & & & \\
& 1& -5& 6 & -16 & 56
\end{array}$
The coefficients of the quotient and remainder alternate signs. thus, $-3$ is a lower bound for the real zeros of $P$.
