Answer
$1$ is an upper bound for the real zeros of $P$
$-2$ 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 $c$.
$P(x)=x^4+2x^3+3x^2+5x-1$; $a=-2$, $b=1$
$\begin{array}{lllll}
\underline{1}| & 1&2 & 3 & 5 & -1& \\
& &1& 3 & 6 & 11\\
\hline & & & & \\
& 1&3 & 6 & 11 & 10
\end{array}$
All of the coefficients of the quotient and remainder are all non-negative values. thus, $1$ is an upper bound for the real zeros of $P$.
$\begin{array}{lllll}
\underline{-2}| &1& 2 & 3 & 5 & -1\\
& & -2&0 & -6& 2\\
\hline & & & & \\
& 1&0& 3 & -1 & 1
\end{array}$
The coefficients of the quotient and remainder alternates signs. thus, $-2$ is a lower bound for the real zeros of $P$.
