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 $c$.
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.
$P(x)=x^4-2x^3+x^2-9x+2$; let's try $3$
$\begin{array}{lllll}
\underline{3}| & 1&-2 & 1 & -9 & 2& \\
& & 3& 3& 12 & 9 \\
\hline & & & & \\
& 1&1 & 4 & 3 & 11
\end{array}$
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 $P$.
Llet's try $-1$
$\begin{array}{lllll}
\underline{-1}| & 1&-2& 1 & -9 & 2 & \\
& &-1& 3&-4 & 13\\
\hline & & & & \\
& 1&-3& 4& -13 & 15
\end{array}$
The coefficients of the quotient and remainder alternate signs, thus, $-1$ is a lower bound for the real zeros of $P$.
