Answer
$2$ is an upper bound for the real zeros of $P$
$-1$ is a lower bound for the real zeros of $P$
Work Step by Step
Upper and Lower bound Theorem states that:
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 $c$.
If the number $0$ occurs as a coefficient of the quotient or remainder in the final line of the division table in either of the above cases, it can be treated as (+) or (−) as needed.
$P(x)=3x^4-5x^3-2x^2+x-1$; $a=-1$, $b=2$
$\begin{array}{lllll}
\underline{2}| & 3&-5 & -2 & 1 & -1& \\
& &6& 2 & 0 & 2\\
\hline & & & & \\
& 3&1 & 0 & 1 & 1
\end{array}$
All of the coefficients of the quotient and remainder are all non-negative values. thus, $2$ is an upper bound for the real zeros of $P$.
$\begin{array}{lllll}
\underline{-1}| &3& -5 & -2 & 1 & -1\\
& & -3&8 & -6& 5\\
\hline & & & & \\
& 3&-8& 6& -5 & 4
\end{array}$
The coefficients of the quotient and remainder alternate signs, thus, $-1$ is a lower bound for the real zeros of $P$.
