Answer
$1$ 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
Upper and Lower bound Theorem states the following:
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.
$P(x)=2x^3+5x^2+x-2$; $a=-3$, $b=1$
$\begin{array}{lllll}
\underline{1}| & 2 & 5 & 1 & -2& \\
& & 2 & 7 & 8\\
\hline & & & & \\
& 2 & 7 & 8 & 6
\end{array}$
All of the coefficients of the quotient and remainder are non-negative values. thus, $1$ is an upper bound for the real zeros of $P$.
$\begin{array}{lllll}
\underline{-3}| & 2 & 5 & 1 & -2\\
& & -6 & 3 & -12\\
\hline & & & & \\
& 2 & -1 & 4 & -14
\end{array}$
The coefficients of the quotient and remainder alternate signs. thus, $-3$ is a lower bound for the real zeros of $P$.
