Answer
$3$ 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 $c$.
$P(x)=2x^4-6x^3+x^2-2x+3$; $a=-1$, $b=3$
$\begin{array}{lllll}
\underline{3}| & 2&-6 & 1 & -2 & 3& \\
& &6& 0 & 3 & 3\\
\hline & & & & \\
& 2&0 & 1 & 1 & 6
\end{array}$
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 $P$.
$\begin{array}{lllll}
\underline{-1}| &2& -6 & 1 & -2 & 3\\
& & -2&8 & -9& 11\\
\hline & & & & \\
& 2&-8& 9& -11 & 14
\end{array}$
The coefficients of the quotient and remainder alternate signs, thus, $-1$ is a lower bound for the real zeros of $P$.
