Answer
$2$ is an upper bound for the real zeros of $P$
$-3$ is a lower bound for the real zeros of $P$ and It is a real zero 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 $c$.
$P(x)=8x^3+10x^2-39x+9$; $a=-3$, $b=2$
$\begin{array}{lllll}
\underline{2}| & 8 & 10 & -39 & 9& \\
& & 16 & 52 & 26\\
\hline & & & & \\
& 8 & 26 & 13 & 35
\end{array}$
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 $P$.
$\begin{array}{lllll}
\underline{-3}| & 8 & 10 & -39 & 9\\
& & -24 & 42& -9\\
\hline & & & & \\
& 8& -14 & 3 & 0
\end{array}$
The coefficients of the quotient and remainder alternates signs. thus, $-3$ is a lower bound for the real zeros of $P$ and It is a real zero of $P$.
