Answer
$1$ 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.
$P(x)=x^5-x^4+1$; let's try $1$:
$\begin{array}{lllll}
\underline{1}| & 1&-1 & 0 & 0 & 0& 1 \\
& & 1& 0& 0 & 0&0 \\
\hline & & & & \\
& 1&0 & 0 & 0 & 0&1
\end{array}$
All of the coefficients of the quotient and remainder are all non-negative values. thus, $1$ is an upper bound for the real zeros of $P$.
Let's try $-1$:
$\begin{array}{lllll}
\underline{-1}| & 1&-1& 0 & 0 & 0& 1\\
& &-1& 2&-2 & 2&-2\\
\hline & & & & \\
& 1&-2& 2& -2 & 2&-1
\end{array}$
The coefficients of the quotient and remainder alternate signs, thus, $-1$ is a lower bound for the real zeros of $P$.
