Answer
$3$ 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 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.
$P(x)=2x^3-3x^2-8x+12$; let's try $3$
$\begin{array}{lllll}
\underline{3}| & 2 & -3 & -8 & 12& \\
& & 6& 9 & 3 \\
\hline & & & & \\
& 2 & 3 & 1 & 15
\end{array}$
All of the coefficients of the quotient and remainder are all non-negative values. thus, $3$ is an upper bound for the real zeros of $P$.
Let's try $-3$
$\begin{array}{lllll}
\underline{-3}| & 2& -3 & -8 & 12 & \\
& & -6&27 & -57\\
\hline & & & & \\
& 2& -9& 19 & -45
\end{array}$
The coefficients of the quotient and remainder alternate signs, thus, $-3$ is a lower bound for the real zeros of $P$.